
Class 

Book L_ 

Copight'N?. 



i^5 



COPYRIGHT DEPOSnV 






•1 



STRENGTH OF MATERIAL 




STRENGTH OF MATERIAL 



AN ELEMENTARY STUDY 

Prepared for the use of Midshipmen at 
the U. S. Naval Academy 



BY 

H. E. SMITH 

Professor of Mathematics, U. S. Navy 



1908 



<i^ 



LIBRARY of CONGRESS 
Two CoDies Received 

WOY 28 1908 

^ Copyriifht tntry^ 
CLASS O^J^U, No. 

COPY a. 



J 



Copyright, 1908, by 

H. E. SMITH, U.S.N. 



U. S. A. 



BALTIMORE, 



^ PEEFACE. 

This book has been prepared for the use of the Midshipmen 
at the U. S. Naval Academy and is- designed to cover a short 
course in the subject taken up in the , Department of Mathe- 
matics and Mechanics preliminary to the work in the Depart- 
ments of Ordnance and G-unnery and of Steam Engineering 
at the xlcademy. 

In arranging the subject matter many of the methods in- 
troduced by officers previously on duty in the Department of 
Mathematics have been employed and the endeavor has been 
to lead the student to the opening point for the professional 
work carried on by the other Departments. 



COXTEISTTS. 

PAGE 

Introduction 9 

Chapteb I. Stress and Strain, Tension and Compression, ... 13 

II. Shearing Force 26 

III. Torsion 38 

IV. Stress due to Bending 46 

V. Combined Stresses 56 

VI. Shearing Stress in Beams 66 

VII. Bending Moments, Curves of Shearing Stress and 

Bending Moments 75 

VIII. Slope and Deflection of Beams 88 

IX. Slope and Deflection (Continued) 97 

X. Continuous Beams 108 

XI. Columns and Struts 118 

XII. Stress on Members of Frames 125 

XIII. Framed Structures 136 

XIV. Framed Structures (Continued) 147 

Miscellaneous Peoblems. Reinforced Concrete Beams, Pois- 

son's Ratio, Stress in Thick Cylinders and 
Guns, Built-up Guns, Stress due to Centrifu- 
gal Force, Bending due to Centrifugal Force. 

Flat Plates 156 



mTRODUCTION. 

Before beginning the stiid}^ of Strength of Material, let us 
see what has been discovered by experimenting with test pieces 
of material and note some of the conclusions arrived at from 
the results obtained in this way. 

Experiment shows us that whenever a force acts on a body 
formed of any substance the dimensions of that body are 
changed. In mechanics all bodies were assumed rigid and 
the results obtained under this assumption were true, for 
mechanics taught us to find the action of one body on another 
or the force transmitted by one body to another, while strength 
of material will teach us the effect in the Ijody itself of a force 
acting upon it. Mechanics showed us that by means of a 
piece of material force could be moved from one point to 
another, and strength of material will show^ us that in trans- 
mitting the force the substance forming the conveyance suffers 
some slight temporary deformation if the force is within 
certain limits; that be3^ond these same limits the substance 
suffers permanent deformation and if the force be great 
enough will be completely ruptured. 

The study of strength of material will include finding the 
safe limits of a force to be transmitted by any particular piece 
of material, finding the deformation caused by transmitting 
any force, and finding the dimensions of a piece of material in 
order that it may safely transmit any particular force. 

With regard to deformation materials differ greatly, for 
example the force which will double the length of a piece of 
rubber will not apparently change the length of a piece of steel 
of the same size, though both of these substances are elastic, 
and each if stretched within limits will return to its orio^inal 



10 Introduction. 

length when the stretching force is removed. The force 
which makes an indentation in a piece of putty will scarcely 
affect a piece of lead, but in this case the indentation made 
will remain in both these substances as they are plastic. 
Obviously, then, we must experiment with the different ma- 
terials and find out some of their physical properties before 
proceeding with a mathematical investigation. 

The materials used in building are elastic and we determine 
their physical properties by experimenting with small pieces 
of them in machines made for the purpose. 

Take steel, for example; small test pieces of different 
shapes are tried under different forces. A pull is applied and 
we find the test pieces stretch; if we apply pressure the test 
pieces are compressed. Having applied all sorts of forces we 
compare our results and find that the stretching or compres- 
sion is always of the same amount if the same value of force 
on unit cube of the steel is used. We also find that up to 
a certain limiting value of the force the material will always 
return to its original shape when the force is removed, but if 
we go beyond this value we find the piece will not return to 
its original shape; in other words it. is not perfectly elastic 
for forces beyond this value. 

If we go through the whole list of materials used in build- 
ing and test each kind in the same way we will find that they 
will all behave in a similar manner, the difference will lie in 
the amount of deformation for any force and the value of the 
limiting force beyond which they are not perfectly elastic. 

Experimenting further we find the value of this limiting 
force for the different materials and having it we can compare 
the various substances as to their usefulness under different 
circumstances. 

If we continue to experiment again and again with a single 
substance we find that we may apply as often as we like a 



IKTEODUCTIOj^. 11 

force less than the limiting one we found, and that the piece 
will alM-ays return to its original shape when the force is 
removed; but when the force used exceeds the limiting one 
found, if by ever so little, and the force is applied and re- 
moved often enough the piece will 'break, though the deforma- 
tion caused by the first application of the force was too small 
to be measured or even noticed. From this fact we see that 
we must never use a piece of material which will have to 
sustain a force w^hich is in the least greater than the limiting 
value found by experimenting. 

Materials differ in other ways. A piece of glass is easily 
broken by a light blow and is therefore called hrittle or 
fragile; wrought iron can be twisted and bent into almost any 
shape without rupture and therefore is called malleable or 
ductile. 

Material which has been melted and cast into desired shapes 
cools quicker at some parts than it does at others, thereby 
setting up within it internal stresses which are irregularly 
distributed. Such castings can be broken by a comparatively 
light blow though they can usually withstand a large pressure. 
These internal stresses can be removed by a process called 
annealing, which consists in heating the body to a red heat 
and allowing it to cool slowly, thus allowing the particles an 
opportunity to rearrange themselves. 

Metals have the peculiarity that if overstrained they harden 
in the vicinity of the overstrain and this hardening goes on 
with time. Thus a bar which is sheared off while cold will 
finally become extremely hard and brittle near the sheared 
end, as will a plate in the neighborhood of cold-punched rivet 
holes. To avoid this bore the rivet holes and saw the bar, or 
if feasible anneal the cold-sheared bar and the plate near the 
cold-punched rivet holes. 

Xow in all practical cases we must of course use material 



12 



Introduction^, 



that will not break^ but in addition we must have material 
which will not change its dimensions to any considerable 
extent under the applied load. 

The experiments by which the constants used in this study 
have been determined were very carefully conducted and are 
the results of many independent efforts on the parts of many 
different scientists. In this work the mathematical investi- 
gation only will be touched upon, the experimental part being 
beyond its scope. 



STRENGTH OF MATERIAL 



CHAPTEK I. 

Tension and Compression — Stress and Strain. 

1. In the study of strength of material we must consider 
two ways of arranging the different pieces used: firsts when 
there is to be motion between the parts^, and second, when the 
parts are to be relatively at rest. In the first case, force is 
transmitted from one piece to another and the combination 
of pieces is called a machine, the study of which involves the 
principles of d3'namics; the second arrangement is called a 
framed structure, or simply structure, and we must employ 
the principles of statics in its investigation. In either 
arrangement, any two parts in contact have a mutual action 
between the touching surfaces, and the effects produced by this 
action depend in great measure upon the way in which it is 
applied. In any case it tends to change the shape or dimen- 
sions of the parts involved and if the force is great enough to 
crush or break them. So for permanence the machine or 
structure must be strong enough m each part to withstand 
any force to which it may be subjected. If then we can find 
the greatest force that any particular piece of material can 
endure without breaking or suffering a permanent change in 
shape we can be sure of its remaining intact for all forces 
within that limit. In the first four chapters we will apply, 
separately, all the different forces to which a piece of material 
can be subjected, and as any piece in our machine or struc- 
ture may have more than one of these forces to sustain at 
any given instant we will, in the fifth chapter, show the effect 
of the combined action of two or more of them. 
1 



14 Strength of Material. 

2. Let "lis first investigate the effect on a piece of material 
of an external force applied to it, and we will choose for our 
investigation a straight rod of uniform cross-section and will 
not consider the force of gravity as acting. We will apply 
to the end of our rod a pull, F (not sufficient to break or per- 
manently change its shape), which, in order that the rod 
remain stationary, will require an equal pull in the opposite 
direction at the other end. These forces tend to tear apart 
the particles of the material, and as the bar remains intact 
the particles must be in a different condition, relative to each 
other, from that in which they were before we applied the 
pull. If instead of a pull we exert a pressure on one end, an 




Fig. 1. 



equal and opposite pressure must be applied to the other end 
to keep the bar in equilibrium, and the particles of the 
material will now tend to crowd together and crush each 
other. In both of these cases we have arranged our forces 
so that the bar does not move and they have been taken small 
enough so as not permanently to change its dimensions. I^ow 
as the length of the bar separates the points of application of 
our forces, there must be some action set up among the par- 
ticles of the bar itself which transmits the force from one 
end to the other, or to a common point of action. Let us now 
imagine a plane passed through the bar perpendicular to its 
axis: in the first case (the pull, Fig. 1) our forces would tend 
to pull apart the two pieces of the bar; in the second (not 
shown) they would press them together, each piece would 



Strength of Material. 



15 



tend to move in the direction of the external force acting on 
it and the amount of this tendency wonld be equal to that 
force so that the total action in the bar between the particles 
on either side of any imaginary section wonld be eqnal but 
opposite to the external forces applied. The action of all the 
particles on the right side of any section would be equal to 
hut opposite to the force on the right end, and of those on the 
left side equal to hut opposite to the force on the left end. 
This must be true in order that the bar remain intact, i. e. in 
equilibrium. 

^iU4 



^rxvttxziQf^. 



/>rKSSu.v^ 



Fig. 2. 



3. Stress. — When any such action as the above is set up 
among the particles of a piece of material that piece is said 
to be under stress, and the external force which causes the 
stress is called the load. A rod under the action of a pull is 
said to be in tension, and the pull is the tensile load. A rod 
to which pressure is applied is said to be under compression, 
and the pressure is called the compressive load. Here- 
after in using the word stress we will mean the amount of 
action between the particles in unit cross-sectional area, and 
will use " total stress ^' for the action over the area of the 
whole cross-section. The stress per unit cross-sectional area 
is sometimes called " intensity of stress ^' and is equal to the 



16 Strength of Material. 

external force^ F, divided by the cross-sectional area, A. The 
forces on one side of the section only are nsed^ so, 

]} (stress) = -^ . 

Of course we wonld get the same result by using the forces 
on the other side of the section^ as the bar must be in equi- 
librium; the}^, however, would act in the opposite direction. 

4. Now let us see what will happen if we gradually increase 
the load F. We know that all materials are more or less 
elastic, so as F is increased the bar will stretch or shorten 
according as F puts it in tension or compression ; and up to a 
certain value of F, the bar will spring back to its original 
length when F is removed. If we experiment carefully with 
any material we will find that there is for it a certain value of 
F, after reaching which, and then having removed F, the bar 
will be found a little longer or a little shorter than it was 
originally; in other words, it will have a '^permanent set." 
This value of F for unit sectional area is called the " limiting 
stress " or " elastic limit " of that material. It is obvious 
that no part of our machine or structure must be subjected to 
a force equal to this, for if it be the piece so used is afterward 
unfit to do its work. 

If we go on increasing F the bar will finally break or crush, 

but at present we are interested only in the elastic limit, the 

stress for which we will call /, and for several materials its 

value will be found in the table at the end of this chapter. 

F 
We know that any stress, p, is equal to^-, so to find the 

tensile or compressive load any piece of material can sustain 

F 

unhurt, we put / =— - or i^ (the limiting load) = fA. This 

is true for all ordinary lengths of material under tension, and 
for short pieces under compression. We will later (Chapter 
XI) consider long pieces under compression, in which the 
stress due to bending must be taken into account. 



Strength of Material. 17 

5. In the preceding article we saw that^, within the elastic 
limit, a piece of material would, when the load was removed, 
return to its original length. After voluminous experiments, 
it has been found that within the elastic limit, the extension 
or compression of a rod under stress varies directly as the 
stress, or stress is equal to a constant multiplied by the ex- 
tension or compression. This is known as Hoolce's Lqw, from 
its discoverer. If we let x denote the total extension of a 

rod of length I, the extension per unit length will be ^^ and 

this extension per unit length is called strain, and we will 
denote it by e; then if p denote the stress per unit area of 
cross-section, 

p = E-j- = Ee, 

E being a constant found by experiment and called the 
modulus of elasticity. Its value for several materials will be 

found in the table at the end of this chapter, £' = ^ . 

6. Work Done on a Piece of Material by a Load. — In all 

cases of overcoming resistance work is done, and we will now 
find how much work is done on a piece of material on which 
a load is acting. By Hookers Law p ^=^ Ee bo that, as we have 
equilibrium within the elastic limit, the resistance of the 
material to the load must equal p per unit area. Before the 
load is applied the resistance is nothing. N'ow let us apply 
the load very slowly. The resistance will increase with the 
load from zero to the equal of the final load. It follows that 
the mean resistance will be equal to one-half of the final load. 
Work is equal to force multiplied by the space through which 
it acts. Due to the load, our rod has stretched through the 
distance x. Hence the work done is equal to one-half the 



18 



Strength of Material. 



final resistance multiplied by x, or if L be the final load the 



work is equal to -^ X ^^ 



W = 



Lx 



The following graphic method may be clearer : Let the 
abscissa represent the extensions^ and the ordinates the loads 
causing them. When x := 0, L =^ Q, and as the load in- 
creases so will the extension in accord with the formula 
y = Ee, giving us a straight line for the load curve. The 
work done will now be represented by the area between the 
load curve, OA, the axis of x, and the ordinate. Ah, represent- 
ing the final load L. 




Fig. 3. 



Work = ^ X -^ = area of triangle OAh. 



The following is the method by calculus : 
L (load) ^pA (Art.4), 

p = E^=Ee (Art 5). 

The load curve then is 

L = ^A = A£' I = AEe. 



Stren^gth of Material. 19 

The work done by any load L acting through a space dx is 
Ldx or dW ^ Ldx, 



but 

T — W.A^ 

r 



EA"^ 



dW = ^ xdx, 

which integrated between the limits for x gives 
EA x^~\ total extension 



It 

wmc 



^=^-^j. 



which as p = — ^ and i = ^A gives Tf = -^' for the work as 

before. 

If our bar be stretched to just within the elastic limit we 
will have 



also 



f^E-^OTX^^ 



f^^ovL^fAj 



hence, from the above, 

W = £^ = -4^ XV (V equals volume of bar) . 

As we have stretched our bar to just within the elastic 
limit, it will return to its original length when we remove the 
load. This load then does the greatest amount of work that 
can be done on a piece of material without injuring it. While 
stretched, the rod has stored in it an amount of energy equal 
to the work done in stretching it to its elastic limit. This 
stored energy is called the resilience of the bar, and the part 

^-^is the modulus of resdience. 
%E 



20 Strength of Material. 

7. The work done by forces quicMy applied is much greater 
than if the force is slowly increased, for, suppose a vertical 
rod having a collar round its lower end is stretched by a 
weight W falling from a height li upon the collar. The work 
done by the falling weight is Wiji -\- x) and this must equal 

the work done on the rod or — — . We have then 

W{}i + x)^^. .'. L = 2W(^ + ^V 

or the slowly applied load which would stretch the rod to the 
same extent as the falling weight would have to be consider- 
ably greater than the weight. 

Again, if we imagine a load L to be instantaneously applied 
and to cause a sUmii equal to x and a load L^^ , slowly applied 
which would cause the same strain, the work done in the first 
case or Lx (force times space) must equal the work done in 

the second case or -^ , Equating we have L^ = 2L or a 
2 

suddenly applied load has twice the effect of a slowly applied 

one. 

Definitions: 

Stress. — ^When due to the load on a body, there is mutual 
action between the particles on either side of a section through 
the body, so that the particles on one side exert a force on 
those on the other side, stress is said to exist in that body, 
and the intensity of the stress is the force per unit area of 
cross-section. Briefly, then, stress is force per unit area of 
cross-section. 

Strain. — ^The ratio of change of length in a body due to 
the load on it, to the original length ; briefly, change of length 
per unit length. 



Strength of Material. 



21 



Modulus of Elasticity. — The stress which would double 
the length of a rod provided Hookers Law held good for that 

extension ; or. it is the ratio of unit stress [ ] to unit str 



/total extension \ 



\area / 



am 



or ratio of stress to strain within Hooke's 



\ original length/' 
Law. 

Elastic Limit. — That stress^ which if exceeded will pro- 
duce a permanent change of length; or the maximum stress 
a material can suffer without being permanently deformed. 

Eesiliexce. — When a bar is loaded to its elastic limit, the 
work done in stretching it is called the resilience of the bar, 
and the modulus of resilience is this work divided by the vol- 
ume of the bar; or, resilience is the capacity of a body to 
resist external work. 

Ultimate Strength is the stress which produces rupture. 

Working Load is the maximum stress to which a piece of 
material will be subjected in actual practice. 



strength of material. 





Average values in pounds per square inch. 


Material. 


Weight 

per 

cubic 

foot. 

Pounds. 


Elastic 

limit. 

/. 


Modulus 

of 
elasticity. 

E. 


Modulus 

of 
elasticity 

for 

shear and 

torsion. 

C. 


Ultimate 
fiber 
stress. 


Limiting 

shearing 

stress. 


Steel 

Iron, cast 

Iron, wrought 
Brass, cast .... 
Brass, drawn.. 
Copper, cast... 
Copper, dr'wn 
Stone, granite 
Timber 


490 

450 
480 

"520 
540 

'ieo 

40 


j 35,000 to 1 
1 50,000 f 
j 6,000 T( 

1 20,000 c r 

25,000 

*2,o6d 

3,000 


29,000,000 

15,000,000 

25,000,000 
9,500,000 
14.500,000 
12,000.000 
15,000,000 
6,000,000 
1,500,000 


10,500,000 

5,000,000 
10,000,000 

5,boo",666 
6,000,666 

1,800,000 
140,000 


110,000 

35,000 

54,000 
20,000 
70,000 
22,000 
65,000 
2,000 
10.000 


50,000 

7,850 

20,160 
4,030 

' 2,896 
■ 1,266 



22 Strength of Material. 

Examples: 

1. A steel bar, 5 ins. long, sectional area -J sq. in., stretches 
.007 in. Tinder a load of 20,000 lbs., and shows no permanent 
elongation when the load is removed. What is the modulus 
of elasticity of the metal ? 

Solution : 
^^^^ 80,000X^5X8 = 38,571,428.571 in.-lbs. 

2. A vertical wrought-iron rod 200 ft. long has to lift sud- 
denly a weight of 2 tons. What is the area of its cross-sec- 
tion if the greatest strain to which wrought iron may be 
subjected is .0005 for unit length. E = 30,000,000. Neglect 
the weight of the rod. 

Solution: Stress = 8960 lbs.^= p = Ee. .*. A =-^ 

A ^ Ee 

= ?^ or A = .597i sq. in. 

30,000,000 X .0005 ^ ^ 

3. Find the area of cross-section in example 2, taking the 
weight of the rod into account. 

Ans. .611 sq. in., or .625 if rod also is suddenly raised. 

4. A steel rod ^ sq. in. in sectional area and 5 ft. long is 
found to have stretched y^-q in. under a load of 1 ton. What 
is the modulus of elasticity of steel. 

Ans. 35,840,000 in.-lbs. 

5. A chain 30 ft. long and sectional area f sq. in, sustains 
a load of 3900 lbs., an additional load of 900 lbs. is suddenly 
applied. Find the resilience at the instant the 900 lbs. is 
applied. E = 25,000,000. 

Ans. 25.992 ft.-lbs. 

6. A steel rod 3J ft. long, 2^ sq. ins. sectional area, reaches 
the elastic limit at 125,000 lbs., with an elongation of .065 in. 
Find the stress and strain at the elastic limit, the modulus of 



Strength of Material. 23 

elasticity, and the modulus of resilience of steel, and express 
each in its proper units. 

Ans. / = 50,000 lbs. per sq. in. ; e = .00166f; E = 
30,000,000 lbs. per sq. in. ; and modulus resilience ^ 41| lbs. 
per sq. in. 

7. A piston rod is 10 ft. long and 7 ins. in diameter. The 
diameter of the C5dinder is 5 ft. 10 ins., and the effective steam 
pressure is 100 lbs. per sq. in. Find the stress produced and 
the total alteration in length of the rod for a complete revolu- 
tion. E = 30,000,000. 

Ans. Change in length .0796 in. 

8. How much work can be done in stretching a composition 

rod 5 ft. long and 2 in. in diameter, without injury, if the 

proof stress of the metal is 2.8 tons per sq. in. ? E =^ 4928 

ton-ins. 

22 
Ans. .15 ton-ins., using tt =: -^. 

9. The proof strain of iron being yq-Vo" > what is the short- 
est length of rod l-J sq. ins. sectional area, which will not take 
a permanent set if subjected to the shock caused by checking 
the weight of 36 lbs. dropped through 10 ft., before beginning 
to strain the rod? E = 30,000,000. 

Ans. 192.31 ins. 

10. Find the shortest length of steel rod, 2 sq. ins. sectional 
area, which will just bear, without injury, the shock caused by 
checking a weight of 60 lbs. which falls through 12 ft. before 
beginning to strain the rod. E = 30,000,000. Modulus of 
resilience = 15 lb. -ft. 

Ans. 24.05 ft. 

11. A brass pump rod is 5 ft. long and 4 ins. in diameter 
and lifts a bucket 28 ins. in diameter, on which is a pressure 
of 6 lbs. per sq. in., in addition to the atmosphere, against a 



24 Strexgth of Material. 

vacimm below the bucket which reduces the atmospheric pres- 
sure to 2 lbs. What is the stress in the rod and the total 
extension per stroke ? E ^^ 9,000,000. 

Ans. Stress 925 lbs. per sq. in., and extension .0061 in. 

12. Assuming a chain twice as strong as the round bar of 
which the links are made, what size chain must be used on a 
20-ton crane with three sheaved blocks if / = 6000 ? 

Ans. Diameter of section of metal .8899 in. 

13. K piston rod is 9 ft. long and 8 ins. in diameter. The 
diameter of the cylinder is 88 ins. and the effective pressure 
is 40 lbs. per sq. in. What is the stress produced and the total 
alteration in length of the rod per revolution? E = 29,- 
000,000. 

Ans. X = .0359 in. 

14. Find the work done in Example 13, and find the resil- 
ience of the rod if / = 12 tons. 

Ans. W = 3.88 in.-tons. R = 52.5 in.-tons. 

15. The stays of a boiler in which the pressure is 245 lbs. 
per sq. in. are spaced 16 ins. apart. What must be their 
diameter if the stress allowed is 18,000 lbs. per sq. in.? 

Ans. 2^ ins. 

16. What is the area of the section of a stone pillar carry- 
ing 5 tons if the stress allowed is 150 lbs. per sq. in.? 

Ans. 75 sq. ins. 

17. What is the length of an iron rod (vertical) which will 
just carry its own weight? / = 9000 lbs. per sq. in. 

Ans. 2700 ft. 

18. The coefficient of expansion of -iron is .0000068 in. per 
degree F. An iron bar 18 ft. long, 1-| ins. in diameter, is 
secured at 400° F., between two walls. What is the pull on 
the walls when the bar has cooled to 300° F. ? 

Ans. 34,862 lbs. 



Strength of Material. 25 

19. Coefficient of expansion of copper is .0000095 in. 
E =^ 17,000,000. A bar of iron is secnred between two bars 
of copper of the same length and section at 60° F. What are 
the stresses in the bars at 200° F. ? 

Ans. In copper 2958, iron 5916 lbs. per sq. in. 

20. A bar of iron is 3 ft. long, 2 ins. in diameter. The 
middle foot is turned down to one in. diameter. Compare 
the resilience with the original bar and with a uniform bar 
of the same weight. 

Ans. (1) 1 to 8; (2) 1 to 6. 



26 



Strength of Material. 



CHAPTEE II. 
Shearin^g. 

8. We will next apply force to produce shearing. In study- 
ing shearing we will use the same sort of rod we used in the 
preceding chapter but will apply our force in a different way. 
To get the proper effect we will slip our rod through holes 
bored in two extra pieces of material as in Fig. 4, and then 
apply opposite and equal pulls to the two extra pieces so that 
the effect on the rod will be to pull one part up and the other 
down. If we pull hard enough our rod will be sliced off 



r 

Fig. 4. 



smoothly as though a plane had been passed through it per- 
pendicular to its axis. It will be sheared off. Practically 
this is as near as we can approximate to pure shearing. In 
theory the two forces, F, should act on either side of a section 
of the rod in parallel planes indefinitely close to each other 
so that their pull would induce no tendency to turning or 
bending; then whether we actually sheared the bar through 
or not, the stress in the bar on one side of the section would 



Strength of Material. 



27 



be the force^ F, divided b}^ the area A, of the section of the 

rp 

bar, oT,q=-j. 

It will be noticed that the direction of the shearing stress 
is parallel to the section, while in tensile or compressive stress 
the direction is normal to the section; for this reason shearing 
is sometimes called tangeniial stress. 

9. Two plates bearing a longitudinal load and held together 
by a riveted joint present a good illustration of shearing. 
The rivet is under almost pure shear if it closely fits the rivet 
holes, and the bearing surfaces of the plates are plane. Fig. 5 
shows a section and plan of a single-riveted lap joint. The 
distance between the centers of the rivet holes is called the 



-0 

6 



Fig. 5. 



-^:^ 



pitch, p, and it is obvious that each rivet supports the load on 

a strip of plate equal in width to the pitch of the rivets. So 

that if F is the force acting along this strip of plate, the 

jp 
shearing stress, q, on the rivet supporting it is equal to -^ 

as before, or as A is ^^ , where d is diameter of the rivet, 



F 

we have for the shearing stress q = — -^ 



If we put the 



28 Strength of Material. 

limiting shearing stress for the material of which our rivet is 
made for q, we can find the diameter of the rivet we must use 

under the conditions, d =^2 kf — . 

\ Tvq 

In working joints, such as pin joints, the action is not pure 

shear, but, due to the clearance necessary for a working fit 

there is some bending. Experiment has proved that the stress 

at the center of the pin of such a joint is -^of that found by 

o 

the above formula. To find the diameter of the pin necessary 
for a joint like that in Fig. 6, we first notice that there are 
two sections of the pin to be sheared. 



/-< 



C 



Fig. 6. 

then, 

r shearing stress ") _ ^ _ i j?L _ 8i^ 

t at center of pin J * * ' * ~ 3 ^ ~ 3 ' 2A~ S^id^ 



or. 






The limiting shearing stress for several materials is given in 
the table at the end of Chapter I. 

10. Usually the value of F for the above is readily found, 
as the thrust on a piston rod, etc. ; but for the force acting on 
riveted joints the ordinary steam boiler will give us a good 
example. Fig. 7 is a section through a boiler designed to 
carry a steam pressure of s pounds per square inch, its radius 
is r, and we wish to find the tangential force at any point A. 



Strength of Material. 



29 



Draw the perpendicular diameters AB and CD, then on a 
ring 1 inch wide there will be a pressure of s pounds on each 
inch of the circumference. Now if we should resolve hori- 
zontally and vertically the pressure on each one of these 
square inches, the sum of all the horizontal components would 
be zero. If, however, we take the sum of all the horizontal 
components on the right half of the ring we will get all the 
forces which act toward the right; of course those to the left 




of the diameter AB will be equal but opposite in direction. 
It will be further noticed that to support these forces we have 
the boiler material at A and also at B and both these points 
support equal amounts, so we can divide the force acting on 
the semicircle by 2 to get that part which acts at A or B; or, 
we can take the sum of the horizontal components of the 
pressure on that part of the ring from C to A which will give 
the same result. Finding this sum is most readily done by 
calculus. Take any point on the circumference and let its 
2 



30 Strength of Material. 

angular distance from C be 6, then rd6 (remembering our 
strip is 1 inch wide) will be an element of area on which the 
pressure is 5 X rdd acting radially and of which the horizontal 
component is srd6 cos 0, and if we integrate this expression 

between the limits and ^ we will get the sum of all these 

components^, or the force H. 

H = sr I ^ cos Odd = sr fsin 6/ 1 2" = sr. ^H 

Having H the force on a strip 1 inch wide the force on a 
strip supported by one of our rivets is found by multiplying 
this by the pitch, or i^ = psr, so that the shearing stress on 
the rivet is 

psr 

The vertical components have no tensile effect at A, but form I 
all the tensile effect at C. 

11. In this connection we can find the required thickness 
of a cylindrical shell, such as a steam pipe, remembering that 
it is under tensile stress not shearing and changing our con- 
stants accordingly. If I is the length considered and t the 
thickness of the plate, the sectional area of the metal will 
be It, and if / is the tensile strength allowed, the resistance 
of the shell will be jit; this must be balanced by the force IF, 
which from Art. 10 is Isr: 

so fit ^ Isr; OT t — --r. 

If we wish the thickness of boiler plates, we must, since boil- 
ers are built of plates with riveted joints, divide our result by 
the efficiency of the joint, strength of joint being equal to the 
strength of the solid metal multiplied by the efficiency of 
the joint. 



Strength of Material. 



31 



12. We have seen how to find the shearing stress on a rivet 
of a single-riveted lap joints, but we must remember we have 
talcen out of our plate a piece of metal equal to the diameter 
of the rivet hole^ so that we have left in the strip of plate, to 
bear the whole force ¥, a section whose area is {p — d)t, and 
the tensile stress on this section must not exceed the tensile 

strensfth of the material, or to balance ; i = tn ; 

° tip — d)^ 

transposing, the pitch for a single-riveted lap joint is 

F 

There is another kind of joint called the butt jomt, where a 
narrow strip of material, or strap, covers the edges of the 
plates and is riveted to both. This joint is called a single 
butt joint, single-riveted (Fig. 8), and is treated exactly as 
a single-riveted lap joint. 



Fig. 8. 



If there are two straps used, one on either side of the plates 
as in Fig. 9, the joint is called a double butt strap joint. 



Fig. 9. 



|| single-riveted, and here it will be noticed we have two sec- 

l' tions of rivet to shear, so our formula becomes q = -—. . 
f 2^ 



ii' 



32 



Strength of Material. 



Now suppose we have two rows of rivets in either a lap joint 
or single bntt joint (where more than one row is nsed the 
rivets are generally staggered as in Fig. 10, shown in the 
plan) and we will have two sections of rivets to shear, and our 




Fig. 10. 



formula will be the same as for the single-riveted double butt 

strap, 

F 

and if we have for these joints n rows of rivets, the formula 
becomes 

_ Z_ • 

^- nA' 

If we use double butt straps we will have double the number 
of sections to shear, so have to divide the above value of F by 
2 ; or, for double butt straps 

^ — 2nA' 



Strength of Material. 



33 



Lap joint or 
single butt strap. 

Smgle-nveted . .5 z= — s^ ov Fz=.—r — 



Double-riveted .q = — ;~^ or i' :^ — ^ — 

2—- 
4 

or F—-^, — 



n rows of rivets.? 



7r(?2 



Double butt strap. 

F _ 2q7Td^ 



•<?2 



4. 



d' 



or F^zqrrd^ 



q — 



or i^ = -^ 



2n 



TTd^ 



13. If we have more than one row of rivets the plate will, 
if it be ruptured, carry away along the outer row of rivets 
in every case; for (see Fig. 11) if it did not, there would be 
one or more rows of rivets to be sheared in addition to the 




Fig. 11. 



carrying away of the plate, and clearly the rupture would 
occur where it would require the least force. Consequently, 
as far as the stress on the plate is concerned, it may always 
be computed as we have it in the final part of Art. 12, using 
the limiting stress allowed for the material; or, 

f(p-d}t = F. 

i At the end of Art. 12 we have formula giving F for the shear- 
ing stress of the material of the rivets and if we equate these 
two values of F we can find the pitch of our rivets for equal 



34 



Strength of Material. 



strength (that is^, rupture will be as likely to occur by shearing 
the rivets as by the plate carrying away), if we know their 
diameter and the thickness of the plates. 
Equating we have 

^' = f(p-d)t. 



d+ 



nq-rrd- 



which gives the pitch where particular plates and rivets are 
to be used. 

In these formula? 

p z= pitch of rivets, 

d = diameter of rivets, 

n = number of rows, 

q = limiting shearing stress for rivets, 

f = limiting tensile stress for plate, 

t = thickness of plate. 

14. When a bar is subjected either to tension or compres- 
sion there is shearing stress along awi/ oblique section. Let 



.7" 



-2> C 

Fig. 12. 

the angle BDC = 6, then the intensity of the force F (acting 

in either direction) alons^ the section DB is ^^-pr , but the 

^ ^ area DB 

area DB equals the area of BC X cosec 6, or if A is the area 

of a right section, the intensity of the force on DB is 

-i . The component of this force resolved alonsr DB is 

A cosec 



Strexgth of Material. 35 



^ cos 6 = ^sin e GosO = — ^ sin (20). This being 



i A cosec e A 'ZA 

Tjl 

a tansrential force is shearinoj stress. -; -resolved per- 

^ "^ A cosec 6 ^ 

pendicular to DB is . ^ sin (9 = — - sin^ 0, the normal 

^ A cosec ^ ^ 

j stress. When ^ = or — the shearing stress is 0, but for an}^ 

other value of (9 it is a finite quantit}^, which proves our propo- 
sition. When ^ = the normal stress is 0, but increases with 

^ to a maximum when (9 = -^ , which is as it should be. 

Examples: 

1. Two wrought- iron plates^ 3 ins. wide by J in. thick are 
lap-jointed by a single rivet, 1 in. in diameter. What will be 
the pull required to break the joint? The tensile strength 
being 18 tons per sq. in. What is the efficiency of the joint? 

Solution: Area (platq) = 3 X i — 1 X i = 1 sq. in. 
.*. Strength ^18 tons. Area of section without rivet hole 
= f sq. in. .'. Strength = | X 18 = 27 tons. Efficiency 

of the joint is -J-f or f = 66f ^. Area of rivet section = ~ . 

If it is of the same material as the plate the force necessary 

to shear it is F = ^' = ^^ ^ f^ = U.13 + tons and the 

joint would break by shearing the rivet. 

2. A cylindrical vessel with hemispherical ends, diameter 
6 ft., is exposed to internal pressure 200 lbs. above the atmos- 
phere. It is constructed of solid steel rings riveted together. 
If / = ; tons per sq. in., how thick must the metal be, and 
what is the longitudinal stress in the metal of the ring joint 
whose section is -^ that of the solid plate ? 



36 Steength of Material. 

Solution : t= ^ = 200 X 6 X 12 _ . 

/ 2 X 7 X 2240 • 

Longitudinal stress = irr^p = yV .27rr^(7 or g = ^ J^^ f =: -^ 

5 X 3 X 12 X 200 . , 
^ = 7 X .459 X 2240 = ' ^^^^^ 

3. What must be the thickness of a copper pipe, f in. in 
diameter, to sustain a pressure of 1350 lbs. per sq. in.? 
/ ^ 950 lbs. per sq. in. 

Ans. .533 in. 

4. A single-riveted lap joint, plate -J in. thick, is under a 
load of 3 tons per sq. in. of plate section. Eivets f in. in 
diameter, pitch IJ ins. What is the shearing stress on the 
rivets and the efficiency of the joint? 

Ans. Shearing stress 3.82 tons. Efficiency 60^. 

5. Two plates, f in. thick, are double-riveted with double 
butt straps, rivets 1 in. in diameter. The shearing strength 
of the rivets is f the tensile strength of the plates. Find the 
pitch and the efficiency of the joint. 

Ans. Pitch ^ ins. Efficiency .7778. 

6. What is the pitch of the rivets in a treble-riveted, double 
butt strap joint between plates ^ in. thick, rivets f in. in 
diameter, if the tensile stress of the plates is limited to 10,000 
lbs. per square in., and the shearing stress of the rivets to 
8000 lbs. per sq. in.? What is the efficiency of the joint? 

Ans. Pitch fff in. Efficiency 83 -f fc. 

7. A cylindrical boiler 8 ft. 4 ins. in diameter is under 100 
lbs. per sq. in. pressure. What must be the thickness of the 
plates that the stress may not exceed 4000 lbs. per sq. in. 

Ans. 1^ ins. 

8. What is the pitch of 1-in. rivets in a double-riveted lap 
joint between -J-in. steel plates? Tensile stress 13.2 tons and 



Strength of Material. 37 

shearing stress 10.5 tons per sq. in. What is the efficiency of 
the joint? 

Ans. Pitch 3^ ins. Efficiency 71 + ^. 

9. The steel plates of a girder are f in. thick, treble-riveted 
with double butt strap, with 1-in. rivets. Shearing stress of 
the rivets is f the tensile strength of the plate. What is the 
pitch ? 

Ans. 6.24 ins. 

10. In a pin joint (Fig. 6) the shearing stress of the pin 
is i the tensile stress of the rod. Compare the diameters. 

Ans. 1 to 8, nearly. 

11. A square bar of steel is under a tensile stress of 4 tons 
and a compressive stress of 2 tons at right angles to its axis. 
What are the shearing and normal stresses on a plane making 

the angle tan"^ -y=- with the axis ? 

Ans. S. 2V^tons. N = 0. 

12. What should be the pitch of 1.25-in. rivets in a treble- 
riveted, double butt strap joint between 1-in. plates, if the 
resistance to shearing is f the resistance to tearing? 

Ans. 6.77 ins. 

' 13. What pressure in a copper pipe, y\ in. in diameter, 

' and .02 in. thick, will stress the copper to 8000 lbs. per sq. in. ? 
Ans. 800 lbs. 



38 Strejs^gth of Material. 



CHAPTER III. 

TORSIOIf. 

15. In Chapters I and II we have seen the effects of tension 
or compression, and of shearing. In this chapter we will fix 
firmly one end of onr rod and twist it by applying a turning 
moment to the free end. This will subject it to torsion, with 
the effect that any right section will be turned about its 
center in an amount depending upon the distance of the 
section from the fixed end.* The stress between the particles 
on either side of any section will be parallel to the section, or 




Fig. 13. 



it is tangential stress. Clearly, then, torsion is a kind of 
shear. If before applying our turning moments, we draw a 
line, AB (Fig. 13), on our test piece, parallel to its axis, it 
will, after the moment is applied, be found to have deformed 
into a helix or screw thread, and the point B will now be at h. 
The angle BAb or cf), is the angle of torsion, and is propor- 
tional to the radius of the rod. The angle at the center, 0, 
is called the angle of twist, and is proportional to the length 

* We might have applied a turning moment to both ends in 
opposite directions, in which case the section at the middle of the 
rod (the fixed end as taken above) would have remained station- 
ary, though it would have been stressed as the others. 



Strexgth of Material. 39 

of the rod. The distance Bh is equal to rO, but it is also equal 
to Z<^; as </) is a very small angle, so that AB and Ah are 
practically equal, 

re = l<\>: orc/,=-^. (1) 

We have, by experiment, the equation q = C<l) for torsion, 
where q is the stress at the surface and C is the modulus of 
distortion or torsion, just as we have p =^ Ee to hold in ten- 
sion or compression. 0, the angle of torsion being a measure 
of the strain. We have then two values of (f>, 

<p = j, and = -^; .-. -j =-^~ ; or ^ ^ ^^. (2) 

It is obvious that for any material, 0, on any section, will be 
constant for any stress. This being true, our equation for 6 
proves that q varies with r^ or if q^ qi , Qo ^ he the stress on the 
surface of rods of the same length of radii r, i\ , r, , then 






(3) 



This is also true by experiment, for radii drawn on any 
section remain straight lines after torsion. 

16. Let us find the stress in a tube so thin that the intensity 
of the stress due to torsion will be practically the same over 
the whole sectional area. If we call this intensity of stress q. 
the mean radius of the tube r, and i the thickness, then the 
area of any section will be l-nri, and the stress on this area 
will be q.^-Kvi, and the moment of this stress about the center 
of the section will be q.^irrt.r. This must balance the mo- 
ment of the force applied to twist the tube, and calling this 
moment T : 

T^2nrHq; or, q=:^. (1) 



40 Strength of Material. 

'Now putting this value of q in equation (2) of Art. 15, which 
is true for either hollow tnbes or solid rods, we have 

for the angle of twist of a thin tube. (Values of C at end of 
Chapter I.,) 

17. Let us now take a tube whose thickness must be con- 
sidered. Let r^ be its external and rg its internal radius 
(Fig. 14), and let q equal the stress at a distance r from the 




Fig. 14. 

center of the tube. Call g^ the maximum stress which is at 
the surface, then from equation (3), Art. 15, 

The element of area at a distance r from the center is 
^irrdr, the stress on it is q.'^irrdr, but from above q = -^ , 
so in terms of the maximum stress the stress on the element is 
^-^.27rrdr and the moment of this stress is ^ 2iri'-dr or 

— ^r^dr, which if integrated between the limits of r which 



Strength of Material. 41 

are r^ and r^ will give us the moment of the resistance offered 
by the tube to the twisting moment applied, or 



-=?'i;«'=i?(v 



2r T 

Knowing T this gives us ^^ — ^ — 



if now r2 = ; or in other words, if the tube is solid, 

^^ = ^' 

18. It is clear that if we give to q-^ the limiting value for 

the material, we can find the diameter of the solid rod that 

will carry a given twisting moment. This formula is used to 

design shafts to transmit a given horsepower. Let a be the 

length of the crank arm, and F the mean force acting on it 

during a revolution; then the mean twisting moment is equal 

to aF, and the work done per revolution is equal to aF.27r 

(care must be taken to use the same units throughout). The 

. ,-, . . 1 X- • H. P. X 33,000 
energy oi the machine per revolution is ^ — - — 

foot-pounds, where H. P. is the indicated horsepower and N 
the number of revolutions per minute. This is also the work 
done per revolution, and if divided by ^tt will give the mean 
twisting moment. The mean twisting moment is less than 
the maximum twisting moment, but the shaft must be de- 
signed to carry the greatest, which is equal to a constant times 
the mean, or T ^ KTm . We put the greatest moment equal 

to -^^ , substitute the limiting stress allowed for q and solve 

for r. We can get the greatest twisting moment if we take 
the force acting on the piston, through the piston and con- 
necting rods to the end of the crank, tvhen the cranTc is per- 
pendicular to the connecting rod, and multiply this by the 
length of the crank, remembering that the force acting along 



42 Strength of Material. 

the connecting rod is the force acting on the piston multiplied 
by the secant of the angle the piston rod makes with the 
connecting rod. 

19. To compare solid and hollow shafts, let the radii of the 
hollow shaft be 7\ and r^ ,■ and that of the solid shaft be r; 
then if T is the resistance offered by the solid shaft and T^ 
that by the hollow shaft 

T, 3r/^i ^2^ < — ^* 



T Ttqr^ 



_2_ 
.J, 2 

' '2 



2 
which gives ns the ratio for shafts of the same material. 
Now if the sectional areas are the same -n-r^ z= 7r{7\- — rj^) 
or r := V^i^ — ^'2^> substituting 

^2 

which shows that the nearer 7\ approaches i\ , or, the thinner 
the metal of the hollow shaft becomes, always retaining the 
same sectional area as the solid, the nearer the ratio of 
strength of hollow to strength of solid would approach oo. 
There is, however, a limit to the thinness of our shaft, as it 
must be able to support its own weight without buckling. 
If we assume that the ratio of the radii of the hollow shaft 
is as 2 to 1, and that the sectional areas are equal, we get 

T 

-^ = 1.44, or the hollow shaft is nearly half again as strong 

as the solid one under these conditions. 

20. Substitute in equation (2) of Art. 15 the value of q 
found in Art. 17 for the solid shaft, and we get the angle of 
twist 

" - -CY 



Strength of Material. 43 

for a solid shaft. If we use q found in that article, for the 
hollow sliaft we get 

_ %Tl 

for the hollow shaft. Now if T is known and (9 can be meas- 
ured, we can find the value of C ; or, if Q is known and Q 
measured we can find T , and thence the horsepower that a 
rotating shaft is transmitting. 

Examples: \ 

1. What must be the diameter of a shaft to transmit a 
twisting moment of 352 ton-ins., the stress allowed being 3^ 
tons per sq. in. ? What H. P. would this shaft transmit at 
108 revolutions per minute, the maximum twisting moment 
being ff of the mean? 

Solution : 

g^^orr3=^- ^-X352X7X2 ^ 

.*. r = 4 ins. ; d ^ S ins. 

^_55 H. P. X 33,000 X 12 
36 • IST X 27r X 2240 
. -TT p _ 352 X 2240 X 108 X 2 X 22 X 36 _ ^^, ^.^ 
••^-^- ^Bp00XT2-X 7'X 55 -884.736. 

2. Find the size of a hollow shaft to replace the preceding, 
exterior diameter to be f the interior. What is the weight of 
metal saved in a steel shaft 60 ft. long? 

Ans. Exterior diameter 8.454 ins. Interior diameter 
5.284. Weight saved 3213 lbs. 

3. Find the diameter of a solid shaft to transmit 9000 H. P. 
at 140 revolutions per minute. The stress allowed being 
10,000 lbs. per sq. in., and the maximum twisting moment f 
the mean. 

Ans. d = 14.5695 ins. 



44 Strength of Material. 

4. Find the size of a hollow steel shaft to replace the above, 
internal diameter being yq of the external. What is the 
saving in weight for 60 ft. of shafting? 

Ans. External diameter = 15.091 ins. Internal diameter 
= 8.4886 ins. Weight saved 8893.5 lbs. 

5. Compare the strength of a solid wronght-iron shaft with 
a hollow steel shaft of the same external diameter. The in- 
ternal diameter of the steel shaft being ^, the external and 
elastic strength of steel f that of iron. 

Ans. As 32 is to 45. 

6. A solid shaft fits exactly inside a hollow shaft of equal 
length. They contain the same amount of material. Com- 
pare their strengths when used separately. 

Ans. As 3 is to V^- 

7. The resistance of a twin-screw vessel at 18 knots is 
44,000 lbs. At 95 revolutions per minute what will be the 
twisting moment on each shaft? What is the H. P.? 

Ans. T = 360 in.-tons. H. P. = 2432. 

8. The pitch of a screw propeller is 14 ft.; the twisting 
moment on the shaft is 120 ton-ins. ; the mean diameter of the 
thrust bearing rings is 15 ins.; coefficient of friction is .05; 
find the thrust and the efficiency of the thrust bearing. 

Ans. Thrust := 4,^ tons. Efficiency of bearing 98f ^. 

9. The angle of torsion of a shaft is not to exceed 1° for 
each 10 ft. of length; what must be its diameter for a twisting 
moment of 16,940 ft.-lbs. 

rC = 10,976,000 in.-lb. units. ^ = y 1 • 

Ans. Diameter = 6 ins. 

10. A solid steel shaft, 10.63 ins. in diameter, is transmit- 
ting 12,000 H. P. at 200 revolutions per minute. If the 
maximum twisting moment is f the mean, what is the maxi- 
mum stress (torsion,) in the shaft? 

Ans. 20,000 lbs. per sq. in. 



Strength of Material. 45 

11. If the modulus of rigidity be 4800 in in.-ton units, 
what is the greatest stress to which a shaft should be subjected 
that the angle of torsion may not exceed 1° for each 10 diame- 
ters of length? 

Ans. 4.2 tons per sq. in. 

12. In changing engines in a ship, the number of revolu- 
tions is increased -J ; H. P. is doubled ; the ratio of maximum 
to mean twisting moment is changed from f to |. ; and the 
strength of material of the shaft is 25 fo greater. What is the 
relative size of the new shaft? 

Ans. The same size. 

13. Find the angle of torsion of a steel tube, 6 ft. long, 
■J in. thick ; mean diameter 12 ins., shearing stress allowed 
4 tons per sq. in. 

Ans. 5°. 7. 



46 



Strength of Material. 



CHAPTEK lA^ 
Bending. 

21. There is one more way in which we can introduce stress 
into our rod and that is by bending it. If we rest our rod on 
supports at its ends and load it at its middle, it will bend into 
a curve. The plane of this curve is called the plane of lend- 
ing {OABO, Fig. 15). The particles on the concave side of 
the rod will tend to crowd together and will be in compres- 





FiG. 15. 



sion, while those on the convex side will tend to pull apart and 
will be in tension; obviously^ there will be some intermediate 
plane, perpendicular to the plane of bending, where there will 
be neither tension nor compression, a plane of no stress. This 
plane is called the neutral surface. The neutral line is the 
intersection of the neutral surface with the plane of bending. 



Strej^gth of Material. 47 

and it gives us the curve into which the rod bends, and is 
therefore called the elastic curve. The intersection of the 
nentral surface with any section of our rod perpendicular to 
its axis is called the neutral axis of that section. In Fig. 15, 
the lines AO and BO meet, being their point of intersec- 
tion. AOB is the plane of bending. CEFD is the neutral 
surface. AB is the neutral line or elastic curve; and if HH 
is any section of the rod perpendicular to AB, then GK is its 
neutral axis. 

22. Now before bending, all sections of our rod are paral- 
lel, but after bending these same sections (assuming that they 




Fig. 16. 

remain plane) will be inclined to each other, being nearer 
together on the concave side, the same distance apart as be- 
fore bending at the neutral surface, and farther apart at the 
convex side. If R is the radius of curvature of the neutral 
line AB (Fig, 16), 0, the angle between the planes of any two 
sections, and y the distance from the neutral plane to any 



48 Strength of Material. 

point in the rod^ then the length of the elastic curve between 
these two sections is BB, and the length between these two 
sections, along a line all points of which are at a distance y 
from the neutral surface is {B ± y)6, or B6 zt yO, so that 
yO is the total change of length between the two sections at a 
distance y from the neutral surface. The change of length 
per unit length is the total change divided by the original 

length, or the strain is -^ = -^ . From Chapter I, Art. h, 
we have the strain due to tension or compression equal to -^. 

Hence -^ = -^ ^ or the stress due to bending at a distance y 

from the neutral plane will be 

E 

The stress varies then as the distance from the neutral surface, 
and must not at the outer surface in the plane of bending 
exceed the elastic limit of the material, or the surface fiber 
stress must be less than /. 

23. We have above our formula for the fiber stress, but do 
not know what value to use for y, as where the neutral axis 
lies is not known. Neither do we know the length of B. We 
will first prove that the neutral axis of any section of a beam 
passes through the center of gravity of the section. We know 
that the stress is maximum compressive at the surface on the 
concave side; decreases with the distance from the neutral 
surface, where it is zero; there changes to tension; and in- 
creases to a maximum at the surface oh the convex side. The 
rod is in equilibrium, a law of which is that the sum of the 
horizontal components of all the forces acting must equal 
zero. Now the loads which cause the bending are all vertical, 
so can have no horizontal components, the only other forces 
are the horizontal stresses at the section, their sum must then 



Strength of Material. 



49 



be equal to zero ; the stresses on the opposite sides being eqnal 
but opposite in direction. Fig. 17 shows the stress on one 
side only, and the section may be of any shape whatever. 



Let AB be the neutral axis. The formula 



^ 
M 



y gives us 



the stress per unit area on any element of area, dA, which is 
throughout at a distance y from the neutral axis. The stress 

Tjl 

on the element is then pdA, OT-^ydA. If we integrate this 




Fig. 17. 



between the limits for our section, we will get the total stress 
on the section, which we know to be equal to zero, or 

j^ /•limit 

-^ ydA=0. (1) 



H 



JSTow the value of the integral, \ydA, divided by the area of 
the section will give us the distance of the center of gravity 
of the section from the line AB, and if this distance were 
zero the line would pass through the center of gravity of the 
section. Now equation (1) equals 0; we know E and R have 

Jr^limit 
' ydA must be equal to 
limit 

zero. Hence the neutral axis always passes through the 
center of gravity of the section. 



50 



Strength of Material. 



24. Determination of E. — Our rod is in equilibrium, one 
of the laws of which is that the sum of the moments of all the 
forces about any axis must equal zero. Being true for any 
axis we will take the neutral axis for the axis of moments. 
We will first find the sum of the moments of the external 
forces. In Fig. 18 let AB be the rod of length I, loaded with 
W pounds in the middle, so that the supporting forces are 
W 



each 



2 



then to the left of the section HH, the moment of 





f 


r 




L 




r 




Tf 


f 




'f 



Fig. 18. 



w w 

the supporting force -^ is -s- ^J and the moment of the forces 
to the right is 



W 



{l—x) — W 



(2-^) 



Wx 
2 



the same value as before. The moment to the left tends to 
turn the part of the beam on the left of the section in the 
direction of motion of the hands of a watch, while that to the 
right tends to turn the right part of the beam in the opposite 
direction; therefore, if we call the first direction positive, the 
other is negative. As in the other cases of equilibrium we 
will use the values found on one side only of the section. In 
this case, the moment of the external forces about the neutral 
axis is the tending moment for that section, and we will 

Wx 



designate it by M, so M ^ 



Considering then the part 



to the left of this section, the moment of the stress in the sec- 



Strength of Material. 



51 



tion must balance the moment of the external forces about the 
neutral axis. Fig. 19 shows enlarged the part of the rod to 
the left of the section. We see that the moment of the ex- 
ternal forces tends to turn this part of the rod in the direction 
indicated by the arrow marked (X), while the moment of the 
stress in the section tends to turn it in the direction of the 
arrow (2), and for equilibrium these moments must be equal. 

As in Art. 23, the stress on the area dA is -^ ndA, and the 



£L^ 



DC- 



ff 



( — 
*< 

A 


J" 


■/-^x.. 


' ^' 




fr 




^IG. 19. 


H 



E 



moment is -^ ijdA . y, which integrated between the limits for 

the sections gives — / y'-dA. The integral \y^dA between 

^*^nmit 
limits is the moment of inertia of the area of the section 
about the neutral axis, and we will designate it by I, so that 

XT 

the moment of the stress in the section is equal to -^ I. This 
must equal M, ^o M- ^l, or R—^.I. In Art. 22 we found 
'P = ^ , SO we now have a general formula for bending, 



I- R 



52 Steexgth of Material. 

In which p = stress at any point at a distance y from the 
neutral surface; 

M = bending moment at any section due to exter- 
nal loads; 

I := moment of inertia of the area of the section 
about the neutral axis; 

R ^ radius of the arc into which the rod is bent; 

E =: modulus of elasticity of the material of the 
rod. 

25. We have now seen the effect of applying forces to our 
rod in all the different ways. If the stresses caused have the 
same line of action, that is, if they act on any section in the 
same or in a diametrically opposite direction, their algebraic 
sum will give us the total stress on the section. For example, 
the stress due to several longitudinal loads is the algebraic 
sum of the loads divided by the area of the section; or, if a 
rod is under a tensile stress and there is also a fiber stress due 
to bending, the maximum stress would be at the convex sur- 
face and equal to the sum of the two stresses, while at the con- 
cave surface, which would be in compression due to the 
bending, the stress would be the difference between the two, 
and would act in the direction of the greater. Another fact 
must be considered in actual practice : When the temperature 
changes, a free rod expands or contracts without stress, but 
if the rod be prevented from expanding or contracting, stress 
is produced. The method of taking the algebraic sum when 
the lines of action are the same is called the principle of 
superposition, which may be stated as follows: The effect 
due to a combination of forces is equal to the sum of the 
effects due to each force taken separately. 

When the stresses caused by our forces Jiave not the same 
line of action, such as combinations of shearing or torsion 
with bending, we must arrive at their maximum effects in 



Stre^^gth of Material. 53 

some other way. In the next chapter we will endeavor to 
show how to find the maximum stress together with its direc- 
tion, which is due to the combined effects of two or more 
stresses which act at right angles. We will then be able to 
combine the effects of any of the forces which we have applied 
separately to our rod in these first four chapters. We will 
find that the maximum stresses due to any of these combina- 
tions are greater than any of the stresses acting singly, and 
that they act on planes which are inclined to those on which 
any of the single stresses act. This will account for the 
apparently erratic manner in which material sometimes car- 
ries away; so in deciding if a single part of a machine or 
structure is strong enough, we must first find to what forces 
it is subjected, and then if it be able to sustain the total 
stresses they induce. 

Examples: 

1. A beam 3 ins. wide by 3 ins. deep is subjected to a 
bending moment of 72 ton-ins. What is the maximum fiber 
stress ? 

Solution : 

p M My 3 

y=T P=T- y=%- 

I=^AV = ^. .■.p= g^Q =24 tons. 

2. An iron I-beam (without weight) of 12-ft. span has 
flanges 4 ins. by 1 in., and web 8 ins. by J in. What is the 
greatest central load it can carry if the stress is limited to 4 
tons per sq. in. ? 

Solution : 

|-=y. M^,,^ |r=Trx3ft.-lbs.=:Prx36in.-lbs.,/=:^. 

^^5 4 _ W X 36 X 3 r=^X^^^-4 015tons 
^5 542 5 X 36 X 3 ~ ^-^^^ ^°'''- 



54 Strejs^gth of Material. 

3. A cast-iron I-beam has a top flange 3 ins. by 1 in. ; bot- 
tom flange 8 ins. by 2 ins. ; web, trapesoidal, ^ in. thick at 
top and 1 in. thick at bottom; total depth of beam 16 ins. 
Find the position of the neutral axis and the ratio of maxi- 
mum tensile to compressive stresses. 

Ans. I^eutral axis 4.81 ins. from bottom. Eatio T to C 
is 3 to 7. 

4. A wooden beam of rectangular cross-section is 15 ft. 
long and 10 ins. wide. If the maximum bending moment is 
16.5 ton-ft., and the allowed stress is J ton per sq. in., what is 
its depth? 

Ans. 15.4 ins. 

5. An I-beam is 25 ft. long, top flange 3 ins. by 2 ins.; 
bottom flange 10 ins. by 3 ins.; web 12 ins. by 1 in.; total 
depth 17 ins. If the stress is limited to 4-| tons per sq. in., 
find the greatest central load it can support in addition to 
its own weight (take weight of beam as 2000 lbs. acting at 
its center). 

Ans. 6.48 tons. 

6. What is the radius of the smallest circle into which a 
rod of iron 2 ins. in diameter may be bent without injury, the 
stress being limited to 4 tons per sq. in. E = 13,000 ton-ins. 

Ans. R = 270 ft. 10 ins. 

7. A spar, 20 ft. long, is supported at the ends and sus- 
tains a maximum bending moment of 3147.5 Ib.-ft. If the 
stress be limited to ^ ton per sq. in., what is the diameter of 
the spar? 

Ans. 7.0025 ins. 

8. What is the diameter of the smallest circle into which a 
J-in. steel wire may be coiled, keeping the stress within 6 tons 
per sq. in. (E for steel wire being 35,840,000 in in.-lb. units). 

Ans. 111^ ft. 



Streis^gth of Material. 55 

9. A rectangular beam 12 ft. long, 3 ins. wide, 9 ins. dee^D, 
is supported at the ends. Stress is limited to 3 tons per sq. 
in. Find the load which can be carried at the center; also 
find the load if the beam lies the fiat way, i. e., 3 ins. deep and 
9 ins. wide. 

Ans. 1st case 3f tons ; 2d case 1^ tons. 

10. Find the breadth and depth of the rectangular beam of 
maximum strength which can be sawed from a log 2 ft. in 
diameter, and compare its resistance to bending with that of 
the largest square beam that can be sawed from the same log. 

Ans. Top is ~-^ ft., dept is - ft. Eesist bending in 

ratio of 1.089 to 1. 

11. A steel I-beam 30 ft. long; flanges 7 ins. by 8 ins.; 
web 24 ins. by .5 in. ; carries 31,900 lbs. at its center. What 
is the maximum fiber stress? 

Ans. 16,000 lbs. per sq. in. 

12. Compare the resistance to bending of a wrought-iron 
I-beam ; flanges 6 ins. by 1 in. ; web 8 ins. by f in., when up- 
right, and when laid on its side. 

Ans. 4.6 to 1. 

13. A round steel rod, 2 ins. in diameter, can only with- 
stand a bending moment of 6 ton-ins. What is the greatest 
length of such a rod which will just carry its own weight when 
supported at the ends ? 

Ans. 28^ ft. 



56 



Strength of Material. 



CHAPTEE Y. 

Combination of Stresses. 

26. In a rod suffering tension and shear let HH (Fig. 20) 
be any section which has normal stress due to the tensile load, 
F, and tangential stress due to shear. Let the square shown 
represent the base of an elementary cube of volume, whose 
height dz, is perpendicular to the plane of the paper, and on 




whose faces dy, dz, are the stresses P and ;S^ as shown. H is 
clear that the stresses P balance each other, but it will be 
noticed that the tangential stresses S form a moment whose 
effort is to turn the cube to the right. Now the forces are 
assumed within the elastic limit and we know the cube to be 
in equilibrium, so there must be an equal and opposite moment 
tending to turn it to the left. In other words, we must have 
tangential stresses equal to S on the faces dxdz, acting in a 
direction such that their moment will turn the cube to the 
left. These latter stresses are called longitudinal shear, and 



Strength of Material. 



57 



because of the resistance offered to it a solid beam will bend 
less, when supported at the ends, than a pile of thin boards of 
the same volume. We may then concede that the shearing 
stresses at any point within a body in a state of stress, are 
equal and act in planes at right angles ivith each other. 

27. We can now find the plane on which the resultant 
stress, due to a number of stresses acting in planes at right 
angles, is a maximum. Fig. 21 is, enlarged, the elementary 




Pig. 21. 



cube of Fig. 20, with the additional stress P^ , acting so as to 
put it in vertical tension. Let AB be any plane making the 
angle a with the direction of the stress F, and let the sum of 
the components of all the stresses on one side of this plane, 
when resolved along and perpendicular to it, be iV and T , as 
shown. The stresses on the other side of AB would give equal 
and opposite components to iV and T (equilibrium). So that 
our cube is now in equilibrium under the stresses shown. Let 
us resolve these stresses horizontally and vertically, then the 
sum of both the horizontal and the vertical components will 



58 Strength of Material. 

be zero. The intensity of the stress P on AB is P sin a, etc. 
(see Art. 14). Eesolving 

i\^ sin a + T cos a — P sin a — 6^ cos a ^ (1) horizontally. 
iV cos a — r sin a — P^ COS a — /S' sin a = (2) vertically. 

Eliminating T between (1) and (2) we get 

N ^= P sin^ a + Pi cos^ a -}- 2S sin a cos a. (3) 

Eedncing 

/ . 2 1 — cos 2a 2 1 + cos 2a 
sm^ a = : COS^ a = — ^— 



and 2 sin a cos a = sin 2a 1 



]V= ^\^' + ^' ^ ^ cos 2a + ;§ sin 2a. (4) 

By the same method we get 

T= ^~^^ sin 2a -\- 8 cos 2a. (5) 

(4) and (5) give ns the stresses along and perpendicular 
to any plane dne to the combined stresses P, P^ and S. Put- 
ting equal to zero the first derivative with regard to N and a 
or T and a in (4) and (5)^ will give us the value of a for 
which iV or P is a maximum or minimum. 

^ = = — ^'T^ -2 sin 2a + 2;^ cos 2a 
da 2 

gives 

tan 2„ =-p-^or a = i tan-^ p-^ [1 -| • (6) 

^=0 = ^~^' ^ cos 2a — 28 sin 2a 
cia 2 

gives 

tan 2« = ^ or a = i tan- ^ -/■ [ + 1 . (,) 



Strexgth of Material. 59 

28. Equation (6) gives us the angle with P of a plane on 
which the normal stress due to the combined load is a maxi- 
mum (the plane 90° from it giving the minimum), and 
equation (7) the angle with P of a plane on which the 
tangential stress due to the combined load is a maximum 
(minimum 90° from it). Obviously, from these equations, 
the two planes are at 45° from each other (tan 2a being in 
one case the negative reciprocal of what it is in the other), 
or the planes of maximum normal and maximum tangential 
stresses lie at angles of 45° with each other. Equation (6) 
shows the maximum and minimum normal stresses to be at 
right angles, and if we substitute the value of 2a found from 
it, in equation (5), we find that on this plane of maximum 
normal stress the tangential stress is zero, or tlie shear is zero 
when the normal stress is a maximum. The normal stresses 
found in equation (6) and acting on planes at right angles 
are called principal stresses and their directions principal 
directions. Principal stresses are always at right angles. If 
we substitute in equations (4) and (5), the values of sin 2a 
and cos 2a, obtained from equations (6) and (7) respectively, 
we will get the maximum value of the normal and tangential 
stresses. 



iV = ^ + ^' ± V4^'+ (P — P^)' (8) 



+ sign max, 
— sign min. 



and T= ± -J V4^' + (P — Pi^ (9) 

The + sign of (8) gives the maximum tension (or compres- 
sion), and the — sign gives the maximum compression (or 
tension) on a plane at right angles, i. e., principal stresses. 
Using the formula of this article we can find the maximum 
value and its direction of any combination of stresses due to 
the loads used in the preceding chapters. 



60 



Strength of Material. 



29. The Stress Ellipse. — Let us assume that at any point 
within a body there are two normal stresses of intensity^ P 
and P-L (Fig. 22) acting at right angles. Then by Art. 14, 
X, the intensity of the stress P on any plane making an angle 
a with the direction of P is P sin a, and y, the intensity of the 
stress P-L on the same plane, is P^ cos a, 



X 

P 



sm a, 



and|- 



COS a. 




Fig. 22. 



squaring and adding we have 






Sin^ a -\- COS^ a =: 1. 



The above is the equation of an ellipse of wliich the semi- 
axes are P and P-^ , and of which x gfnd y, the coordinates of 
any point {C, Fig. 22), represent respectively the intensity 
of the stresses P and P^ on a plane making the angle a with 
P. The radius vector, OC, represents on the same scale, the 
amount and direction of the resultant intensity of stress on 
the plane AB. The equation of this ellipse in terms of the 



Strength of Material. 



61 



eccentric angle cf>, is a; = P cos <t>, and y = P-^ sin <^. Now 
the tangent of the angle 0, which the resultant stress makes 
with the direction of P is 

-Pi sip _ A cos « _ A ,._ , _ P, 

P COS</) 



tan ^ = ^ = 



X 



Pi cos a P 

-^-^ = -^ tan d) = 

P sm a P ^ 



cot a. 



P^ the resultant stress, is equal to ^/x^ -\- y^, and the angle 
the resultant stress makes with the plane AB is {6-\-a). 
Had there been another normal stress at right angles to the 
plane of P and P^ , the locus of the end of the resultant, C, 
would have been the surface of an ellipsoid. In this case it 
is probably as easy to find the resultant of two of the stresses 
and then get the resultant of the third with it. This ellipse 
is convenient to find the direction and amount of the result- 
ant of the principal stresses, or the combinations of other nor- 




FiG. 23. 



mal stresses, which may be done graphically as follows: 
Draw the axes OX and OY (Fig. 23) and lay off to the same 
4 



62 Strength of Material. 

scale P on OX, and P^ on OY. With these distances for 
radii, describe two concentric circles with centers at 0. Draw 
the line AB, making the given angle a with P and draw OB 
perpendicular to it. Where OD cnts into the circle of radius 
P, drop a perpendicular on OX, and where it cuts the circle 
of radius P^ , draw FG parallel to OX. C, the intersection 
of these last two lines is a point on the ellipse. CE repre- 
sents the stress due to P resolved on the plane AB, and OE 
represents the stress due to P^ , resolved on the plane AB; 
while OC represents the resultant stress on AB due to both, 
and the angle COB is the angle it makes with the plane AB. 

30. Now any condition of stress at any point within a body 
may always be reduced (x4.rt. 14) to three simple stresses 
acting on planes at right angles. By means of the formula 
of Arts. 27 and 28, we can always calculate the value and 
direction of the resultant stress on any plane due to these 
three simple stresses, and by means of the stress ellipse of 
Art. 29, we can calculate or determine graphically the amount 
and direction of the resultant of two simple normal stresses 
on any given plane. For example, if we find a piece of 
material under tensile stress combined with torsion, we can 
substitute for P in equation (8) the tensile stress, and for S 
the stress at the surface (maximum) due to torsion; and, 
unless there is another stress at right angles to the plane of 
these two, P^ is zero. The value of N given by these substi- 
tutions is the maximum normal stress, and the plane on which 
it acts will be found by making the same substitutions in 
equation (6), remembering that a is the angle the plane 
makes with the direction of the tensile stress P. In case of 
bending, the maximum fiber stress should be used for P. 
We must always use the maximum values, for the piece of 
material must be strong enough to sustain the greatest 
stresses to which it will be subjected. 



Strength of Material. 63 

Example d, at the end of this chapter, offers a good illus- 
tration of the nse of the stress ellipse. 

Examples: 

1. The shaft of a vessel, 15 ins. in diameter, is subject to a 
twisting moment of 100 ft. -tons, and a bending moment of 
20 ft.-tons, also the thrust of the screw is 16 tons. Find the 
maximnm stresses on the shaft. 

Soliition : 

or 

N(N— P) = S' and T' = S' -}- ^ . 

4 
Jfv_ 64 . P_ 64 , 

,^ . ^ ^ Fa (twisting moment) 64 X 20 
q (torsion) = S= ^, '- = ^^, . 

16" 

/ 64(l + 8) \ /64^ Y _. r5.93-htons. 

^ ^-^ 7:cl' I {r.d'^^^l- • • ^^ - I 2.2626 + tons. 

^, /^ 64 X 20 V , (64X_9Y ^ . ^.^^ , ^ 

2. A tube, 12 ins. mean diameter and J in. thick is acted 
on by a thrust of 20 tons and a twisting moment of 25 ft.-tons. 
What are the maximum stresses? and their angles? 

Ans. Greater 3.24 tons. 39^°. 

3. A rivet is under shearing stress of 4 tons per sq. in. and 
tensile stress, due to contraction, of 3 tons per sq. in. What 
are the maximum stresses ? 

Ans. Greater 5.8 tons. 



04: Strength of Material. 

4. The thrust of a screw is 20 tons; the shaft diameter 14 
ins., has a twisting moment of 100 ton-ft., and a bending 
moment of 25 ton-ft. Find the maximum stress and compare 
it with what it would have been without the bending moment 
or thrust. 

Ans. Greater 2.9 tons. Eatio 1.32 to 1. 

5. At a point within a solid in a state of stress the princi- 
pal stresses are tension of 255 and 171 lbs. Find the amount 
and direction of the stress on a plane making an angle of 27° 
with the 255 lbs. stress. 

Ans. 191.35 lbs: 79° 46' 20''. 

6. A beam is under a 300-lb. tensile stress and a 100-lb. 
shearing stress. Find the normal and tangential stresses on 
a plane making an angle of 30° with the tensile stress. 

Ans. I^ormal 161.5; tangential 179.8 lbs. per sq. in. 

7. Find the principal stresses and their directions for a 
point in a beam which is under tensile stress of 400 lbs. and 
shearing stress of 250 lbs. 

Ans. Normal maximum 520; minimum — 120; X = 
—25° 40' 12"; or 64° 19' 48". 

8. Find the maximum and minimum values of the shear 
in example 7 and their directions. 

Ans. 320 lbs. per sq. in. Z = 19° 19' 48"; 109° 19' 48". 

9. A bolt, 1 in. in diameter, is under a tension of 5000 lbs. 
and a shearing force of 3000 lbs. Find the maximum stresses 
and their directions. 

Ans. N = 8155 ; T = 4970 lbs. per sq. in. Angles are 
N 64° 53', and T 19° 53' with axis of bolt. 

10. Find the maximum stress in a shaft 3 in. in diameter 
and 12 ft. between bearings, which transmits 40 H. P. at 120 
revolutions per minute, and has a weight of 800 lbs. half way 
between bearings. The shearing stress due to above arrange- 
ment is 4000 lbs. per sq. in. 

Ans. N = 7600 lbs. and T = 4900 lbs. per sq. in. 



Strength of Material. 65 

11. "What is the diameter of a steel shaft to transmit 90 
H. P. at 250 revolutions per minute, the distance between 
bearings to be 8 ft. and a load of 480 lbs. to be carried half 
way between bearings. The allowable maximum stresses 
being 7000 lbs. for N and 5000 lbs. for T (shearing force due 
to middle load 240). 

Ans. J = 2.8 + in. (about 3 ins.) . 

12. A steel bar of rectangular section, 18 ft. long, 1 in. 
thick, and 8 ins. deep, is under tension of 80,000 lbs., and a 
bending moment of 13,230 Ib.-ins. What is the maximum 
stress in the bar? 

Ans. 10,943 lbs. per sq. in. 

13. A wooden beam 8 ft. long, 9 ins. deep, and 10 ins. wide, 
is under compression of 40,000 lbs., and a bending moment of 
48,000 Ib.-ins. What is the maximum stress? 

Ans. 815 lbs. per sq. in. 






66 Steexgth of Material. 



CHAPTEK YI. 

Shearing Stress in Beams. 

31. In the preceding chapters we have made use of the 
simplest kinds of loads. In tension and compression we can 
vary the intensity of the stress only by increasing or dimin- 
ishing the load as it has already been applied; in torsion also 
we can only change the value of the twisting moment by vary- 
ing the amounts of the forces forming it or by changing the 
length of the arm; but to get the stresses due to bending and 
shearing we can apply loads in a number of different ways. 
We have also been neglecting the action of the force of 
gravity on our rod, which is of importance, as in the case of 
large shafts or beams whose bearings or supports must be 
spaced with reference to their weights; beside the supporting 
forces for a beam loaded in any way must be known in order 
to find the shearing and bending stresses. 

32. Obviously, as we have assumed in Chapter IV, the 
supporting forces for a weightless, horizontal beam, loaded 

with a weight, W, exactly at its middle, are each — - , or half 

the load; if, however, we take into consideration the weight 
of the beam and put several loads upon it at different places, 
the conditions of equilibrium require that the sum of the 
moments of all the forces acting on the beam, about any axis 
must be zero. All the forces acting are the loads, the weight 
of the beam and the supporting forces, all acting in the same 
plane.* If we take moments about an axis through one of 

* If the forces do not all act in the same plane those acting in 
one plane may be considered separately, and the resultant of the 



I 



Strength of Material. 



()7 



the supports, the moment of that supporting force will be 
zero, and the other supporting force, being now the only un- 
known quantit}^, is readily found by equating to zero the 
algebraic sum of the moments of all the forces; for example, 
it is required to find the supporting forces of a beam 20 ft. 
long, weighing 10 lbs. per ft., and supported at the ends, 
carrying 150 lbs. 6 ft. from the left end; 300 lbs. 11 ft. 
from the left end; and 750 lbs. 16 ft. from the left end (see 
Fig. 24). The weight of the beam is 200 lbs. and acts at its 
center of gravity, which in this case is at the middle of the 
beam. If we take moments about an axis through the left 




ZO 



Fig. 24. 



end of the beam, we eliminate the supporting force P, whose 
moment about this axis is zero ; our equation for equilibrium 
will then be 

200X10+150X6+300X11+750X16—^X20=0, 

from which § = 910, and P is obviously the difference be- 
tween the sum of the weights and §, or P = 490 lbs. In the 
above the weights act at the points indicated and are known as 
concentrated loads. If the loading is continuous the effect 
is considered as acting at the center of gravity^ as it has been 
taken for the weight of the beam in the above. 

several supporting forces in the different planes found. The sup- 
ports, however, must be strong enough in the directions of each 
of these planes to sustain the forces acting in those planes. 



68 



Stre^-gth op Material. 



33. Having the supporting forces^, we can now find the 
shearing stress on an^^ transverse section of a beam. Eefer- 
ring to Fig. 25, and at first considering the beam weightless, 
if HH is any section between the supporting force P and the 
first load W-^ , there would clearly be a tendency for the two 
parts of the beam to move as shown, the left part remaining 
stationary and the right part being pushed down by the 
weights Tfi , W2 , and TF3; in other words there is in any 
loaded beam tangential or shearing stress along any transverse 
section. The magnitude of this stress (see Chap. II) is 
equal to the load on one side of the section, divided by the 
area of the section. In the beam of Fig. 25, the only load on 




Fig. 25. 

the left of the section HH, is the supporting force P, there- 
fore, the shearing stress on any section between P and the 
P 



first load TFi , is 



A 



Had we taken our section between 



Tfi and Wo we would have had two loads on the left of the 
section and acting in opposite directions, so that the shear- 

P— W 



ino- stress would have been 



and so on across the 



beam, showing a drop as each load is passed. We will assume 
as positive the conditions as in Fig. 25, the tendency being 



* In Chapter II we took our shearing forces indefinitely near to- 
gether to get 'pure shear, that is, shear without bending. Shear 
and bending are closely connected, so that it is difficult to produce 
the one without the other. 



Strength of Material. 



69 



for the part of tlie beam to the right of the section to move 
down; then let iis construct a curve of shearing force using 
for ordinates the values as found above. For example, Fig. 
26 is a beam loaded as shown. Taking the left end as origin, 
the total shearing force at that end is equal to the support- 
ing force there, or S. F. = P = 29 lbs., which it remains 
until, as ^ve move the section to the right, we get to the first 




P=*? 











s 










T 








2.J 

1 


C 






2) 








If 






^ 


>b 




















-/ 




F 


/^ 












t 














1 

-3/ 

1 














0, 




^ 



H 



Fig. 26. 



load, 10 lbs., which acts in the opposite direction to the sup- 
porting force; so, consequently, as we pass this load the total 
shearing stress drops to 19 lbs., where it remains until we 
come to the 20-lb. load. Here it drops to — 1 lb. At this 
point the tendency for the right part of the beam to move 
down ceases, as the sum of the loads on the left side of a 
section from this point on will be greater than the supporting 
force at the left end. Passing the 30-lb. load, the shearing 
stress again drops, this time to — 31 lbs. and the curve of 
shearing force will be the series of steps ABCDEFGH. 



70 



Strength of Material. 



These values, with their signs, are for the left side of the 
section; on the right side the stresses are equal to those 
found but opposite in direction. Had we used the right end 
for origin and the left side down as the positive direction, our 
curve would have been a series of steps down to the left, with 
numerical values the same as before. 

34. If we take the weight of the beam or, what is the same 
thing, consider the beam in the preceding article to be uni- 
formly loaded all along its length, the value of the shearing 




Fig. 27. 



force will change at each point as we move our section to the 
right. Let Fig. 27 represent a beam so loaded; then if the 
ordinate of the line AB represents the load on unit length of 
beam, AB will be the load curve, being drawn below the beam 
or having negative ordinates because the load acts down. 
The supporting forces are now each half the total load, and at 
the left end the shearing force equals the supporting force 
P; but as we move the section to the right we reduce the 
shearing force by the amount of the load between the section 
and the supporting force, the S. F. being equal to P — wx, 
where w is the load per unit length, and x the distance from 



Stkength of Material. 



71 



the left end. The shearing curve will therefore drop steadily, 
as in the figure. The S. F. curve being CD. If we have in 
addition to the uniform load several concentrated loads, the 
curve of shear will be as in Fig. 28, with a drop at each con- 
centrated load. If drawn to scale, these curves will a^ive the 
shearing force at any point of a beam by measuring the ordi- 
nate at that point. As a general rule, then, the shearing 




Fig. 28. 



force at any transverse section of a loaded beam is equal to 
the algebraic sum of all the loads acting on one side of that 
section. 

35. Hereafter we will use the left end of the beam for the 
origin in the equations for curves of any kind. With con- 
tinuous loads, L, the load at any point will vary in some 
way with the distance from the origin; for example, a beam 
" a " ft. long carries a load which uniformly increases from 
zero at the origin to w lbs. per ft.-run at the right end. Here 
the load per ft.-run at a distance x from the origin would be 



72 Stk:en"gth of Material. 

, found from the ratio — = — and the equation of the 

a X a ^ 

load curve (w acting downward) would be L = — — . For a 

ci 

uniform load of w lbs. per ft.-run, L ^ — w. The equa- 
tions for load curves are easily found and are very useful, for 
let us notice the relation between the loads and the shearing 
force at any section. As a general formula for concentrated 
loads, we found the S. F. at any section to be equal to 
P — %W. With a beam loaded with any continuous load, if L 
be the load per unit length, then Ldx is the load on any 
elementary length, dx, of the beam and the total load from 
the origin to a section at any distance, x, from the origin is 

X 

Ldx, the value of L being given by the equation to the load 

curve. 'Now the value of this integral is the algebraic sum 
of all the loads to the left of the section, and that is also our 
definition of shearing force. As a general rule, then, for any 
continuous load the shearing force is 



X 



S. F. =J^ Ldx. 

In using the formula we must remember that the constant of 
integration is the value at the origin of the quantity, repre- 
sented by the integral, in this case that is the value of the 
supporting force P. 

For concentrated loads we must still use S. F. = P — '^W, 
and if a continuously loaded beam supports also a number of 
concentrated loads we must find the shearing forces sepa- 
rately, and get the total shearing force by algebraically adding 
the results (principle of superposition). In this latter case 
also the supporting forces for both the continuous and con- 
centrated loads must be found and used separately. 



Strength of Material. 73 

Examples: 

1. A beam 15 ft. long is supported at the ends and carries 
4 tons 5 ft. from the left end, 1 ton 8 ft. from the left end, 
and -J ton 10 ft. from the left end. Find the supporting 
forces and draw the shearing cnrve. 

Ans. P = 3.3 tons. Q = 2.2 tons. 

2. A spar 20 ft. long is supported at the ends and loaded 
with 500 lbs., 4 ft; 250 lbs., 9 ft; and 900 lbs., 18 ft. from 
the left end. Find the supporting forces and draw the shear- 
ing curve. 

Ans. P — 627.5 lbs. Q = 1022.5 lbs. 

• 3. A plank 16 ft. long is laid across a ditch and a man 
weighing 192 lbs. walks across it. Find the shearing curve 
when he is 3 and when he is 8 ft. from the left end. 

4. A weight of 384 lbs. is placed 5 ft. from the left end of 
the above plank. What is the shearing force at a point 6 J ft. 
from the left end when the man is 8 ft. from there? 

Ans. —24 lbs. 

5. A beam 18 ft. long has a uniform load of 50 lbs. per ft 
run. Draw curve of shearing force and give value at points 
7 ft. and 13 ft. from the left end. 

Ans. At 7 ft. 100 lbs. ; at 13 ft. —200 lbs. 

6. On the beam of example 5, weights of 300 lbs. and 500 
lbs. are placed 6 ft. and 12 ft. respectively from the left end, 
in addition to its uniform load. Draw S. F. curve and give 
value of S. F. 8 ft. from left end. 

Ans. 116| lbs. 

7. A beam 5 ft. long has its left end fixed in a wall and 
supports a load of 1000 lbs. at its free end. What is the 
S. F. ? and draw curve. 

8. The beam of example 7 has a distributed load of 100 
lbs. per ft.-run. What is the maximum S. F. ? and draw the 
curve. 



74 Steength of Material. 

9. An oak beam 15 ft. long and 1 ft. square floats in sea 
water. It is loaded at the center with a weight which will 
jnst immerse it wholly. Draw curve of S. F., and give maxi- 
mum value. 35 cu. ft. of sea water weighs 1 ton; 1 cu. ft. 
of oak weighs 48 lbs. 

Ans. Maximum S. F. = 120 lbs. 

10. A pine beam 20 ft. long and 1 ft. square floats in sea J 
water, and is loaded at the middle with a weight which will t 
just immerse it. Draw a curve of S. F. and give value 5 ft. 
from left end. i 

Ans. S. F. 5 ft. from left end = 125 lbs. 1 

11. A beam 20 ft. long, supported at the ends, carries a 
load which uniformly increases from at the left end to 50 
lbs. per ft.-run at the right. Draw the curve of S. F., and 
find its maximum value, also find the point of the beam where 
its value is zero. 

Ans. Maximum value = 333-J lbs. Value zero where 
X = 11.5 + ft. 

12. The buoyancy of an object floating in the water is 
at the ends and increases uniformly to the center, while its 
weight is at the center and increases uniformly to the end. 
Draw curve of S. F., and give maximum value. 

W 

Ans. Maximum S. F. — -^. 

4 



Stkexgth of Material. 



75 



CHAPTER YIL 
Curves of Bexdixg Momexts and Shearing Force. 

36. In Chapter II it has been shown that when a beam is 
loaded there are horizontal stresses set np on any transverse 
section, these stresses being compressive on one side and 
tensile on the other side of the nentral plane. Art. 14 of the 
same chapter shows that the moment abont the nentral axis 
of the external forces on one side of any section mnst be equal 




to the moment about the neutral axis of the stresses on the 
same side of the section. Definition: The 'bending moment 
at any section is the moment about the neutral axis of that 
section of all the external forces on one side of that section. 
By definition then the bending moment at the section HH 
of a beam loaded as in Fig. 29 would be 

M = Px—W^{x — x^) —W^{x — x^), 

or finding the line of action of the resultant of all the forces 
Fi , F, . etc., and calling its distance from the section " d '' 

M = Px — ^W.d, 
which is true for any kind of load. 

If now we consider the increment AJ/ which the bending 
moment receives if we take our section a distance Aa: to the 



76 Strength of Material. 

right, the bending moment about the neutral axis of the new 
section will be 

M + AM = P(x + Ax)— :^W(d + Ax) 

= Px — ^W.d-{-Ax{P — ^W), 

but Px — ^Wd is the original bending moment equal to M, 
and (P — ^W) is our formula for the shearing force (F) 
at any section, so 

M + AM = M H- FAX, or AM = FAx, 

and passing to the limit dM = F . dx; integrating, we have 
for a general formula for bending moment 

If = jp . fc 

This equation is true for any kind of load, but care must be 
taken to use the correct constant of integration which is the 
value of M at the origin. This for beams free at the origin 
is zero, but if a beam is fixed (prevented from moving in any 
way) at the origin, there is a bending moment there. We 
will devote the rest of this chapter to some examples of shear- 
ing force and bending moments in beams loaded and sup- 
ported in different ways. 

37. A beam supported at the ends, is loaded as shown in 
Fig. 30. Find the curves of shearing force and bending 
moment. By definition the S. F. := P — ^W, by which we 
get the curve of S. F. to be ABCDEF (Fig. 30, a). Consid- 
ering each load separately, the supporting forces for the 25-lb. 
load are P = 17.5, Q = 7.5, and the S. F. curve is ABCD 
(Fig. 30, &). The 35-lb. load gives supporting forces P = 7, 
Q = 28, and the S. F. curve is ABCD (Fig. 30, c). If we 
add algebraically the ordinates given by these curves at any 
point distant x from the origin we will get the ordinates for 
the S. F. curve for that point given in Fig. 30, a. 



:lS'^ 



36'^ 





78 Strength of Material. 

Obtaining the ordinates by definition, the curve of B. M. 
for the beam with these loads is given by OGHX in Fig. 30, a. 

Let us get separately the curves of B. M. due to each load. 
For the 25-lb, load we get by definition the curve in Fig. 
30, d, the maximum B. M. being directly under the load and 
the curve being positive at all points. For the 35-11). load we 
get the curve in Fig. 30, e, this curve also being positive at all 
points. If we add together the ordinates given by these 
curves at any point distant x from the origin we will get the 
ordinate of the curve of B. M. for that point as given in 
Fig. 30, a. Now by definition, the bending moment at the 
middle of the beam, for example, is 

24.5 X 5 — 25 X 2 = 72.5 Ib.-ft. 

and by the formula \F . dx \i i^ — .5X5^ — 2.5, which is 
evidently incorrect, but is due to the fact that having taken 
the two concentrated loads together, the shearing force of one 
being positive and the other negative, we have not multi- 
plied the total shearing force by x. The moments of the two 
shearing forces cause bending in the same direction, therefore, 
if we neglect the negative sign (which we have assumed to 
indicate direction only) the shearing force at this j^oint will 
be 7 + 7.5 ^ 14.5 which multiplied by 5 gives 72.5 as be- 
fore. With many loads to get the shearing force separately 
would be a tedious operation, so it is better to mal-e it a rule 
to get the S. F. and B. M. for concentrated loads from the 
definition. We will find no trouble with beams having con- 
tinuous loads unless there are concentrated loads in addition, 
in which case we get the curves due to the continuous loads 
by formula, and those due to the concentrated loads by defi- 
nition, and apply the principle of superposition. 

38. A beam 10 ft. long supported at the ends is loaded with 
a uniform load of 25 lbs. per ft.-run. Find the curves of 



Strength of Material. 



79 



S. F. and B. M. In this case, a continuous load, onr formula 
is very convenient. L = — lo =: — 25 lbs. 

F = SLdx = — 35a: +C, C being the S. F. at the origin 
is equal to P = 125, lbs. .'. F = 125 — 25a:. (1) 

(1) is the equation to the curve of S. F. This being an 
equation of the first degree is a straight line and plots as the 
line AB of Fig. 31. The bending moment is 
M = S^dx = j (125 — 25a;) dx, 




Fig. 31. 



or, 



M 



125a: 



35a:^ 



+ [(7 = 0, 



(^) 



B. M. is zero at the origin. 
(2) is the equation of the curve of B. M., and being of the 
second degree is a conic (parabola) and plots as OCX of Fig. 
31. The maximum B. M. is at the middle of the beam where 
the S. F. is zero. To get values for either S. F. or B. M. at 
any section of the beam, substitute for x the distance of the 
section from the origin in equations (1) or (2) respectively. 



80 



Strength of Material. 



39. A beam 5 ft. long, fixed at the left end, with the right 
end unsupported, carries a weight of 50 lbs. on the right end. 
Find the curves of S. F. and B. M. 

In this case the left end supports the whole load, so P = 50, 
and the shearing force by definition (concentrated loading) 
is F ^ 50 lbs. The curve being the straight line A. B. of 
Fig. 32. The bending moment by definition is If = Px, 
which being an equation of the first degree is a straight line, 
the ordinates varying from zero to 200 Ib.-ft. This would 
give us a line inclined upward from 0, but we hnow the great- 




FiG. 32. 



est B. M. is at the origin. N"ow notice that the curvature of 
this beam is just the opposite of that of a beam supported at 
the ends, the center of curvature being below the beam in this 
case, while it is above a loaded beam supported at the ends. 
In fact if we turn Fig. 32 " upside down " we will have just 
one-half of the beam supported at the ends and loaded with 
2P in the middle, W being one of the supporting forces; so 
this kind of bending is called negatvve, and instead of the 
curve of B. M. inclining upward from 0, it inclines down- 
ward from X as in the figure. The curve plots directly if we 
take the origin at X and move the section to the left, for this 
arrangement is the same as a beam twice as long supported 
in the middle and loaded with 50 lbs. at both ends. 



Strength of Material. 



81 



40. A beam 5 ft. long fixed at the left end, with the right 
end unsupported carries a nniform load of 25 lbs. per ft. rnn. 
Find the curves of S. F. and B. M. 

A continuous load; so we will use the formula 

F— \Ldx — — 2hx^C. 

C being the S. F. at the origin is equal to P := 125 lbs. 

.-. P = 125 — 25a;. (1) 

(1) being an equation of the first degree is a straight line 
and plots as AX in Fig. 33. The bending moment is 




Fig. 33. 



M— ^Fdx= i{125 — 25x)dx = 125a: 



25a;2 



+ a,. 



Here C^ is not zero, but we know there is no bending moment 
at the right end of the beam so we substitute ilf = and 
x^= 5, and solve for C^ 



= 125 X 5 — ^L^_15 + c^ , 



from which G-^ 
get 



312.5, substituting this value of C^ we 

2>^x' 



M — 125a; 



312.5 



(3) 



82 



Strength of Material. 



for the equation of the curve of B. M., which being of the 
second degree is a conic (parabola) and plots as XC in 
Fig. 33. 

This equation gives a maximum negative value of B. M. 
at the origin, where we know it should be. 

41. A beam 10 ft. long, supported at the ends, is loaded 
with a uniform load of 25 lbs. per ft.-run, and a concentrated 




Fig. 34 



Find the curves of 



load of 100 lbs. 4 ft. from the left end, 
S. F. and B. M. 

For the uniform load (Art. 38) the curve of S. F. is AB 
(Fig. 34). For B. M. the curve is OIX. 

For the concentrated load the S. F. curve is CDEF, and 
the curve of B. M. is OJX. 

Adding algebraically the respective ordinates, we get the 
S. F. curve for both loads to be GHMN, and for B. M. the 
curve of both loads is OLX. 



Strength of Material. 



83 



42. A beam 10 ft. long is supported at the ends and has a 
load uniformly increasing from zero at the left end to 100 
lbs. per ft.-rnn at the right end. Find curves of S. F. and 
B. M. Here (see Art. 35) 



IOO2; 



10 
S. F. = SLdx = 



— 10a;, 



lOxdx ^ 



lOa;'^ 



-Fa 



where C being the S. F. at the origin is equal to P. We can 
find P by the method used in the example at the end of Art. 
32, as follows : In Fig. 35 OL is the load curve, the center 




Fig. 35. 



of gravity of the load then acts at | the length of the beam 
from 0, and the total load is 100 X ^ = 500 lbs., therefore, 

taking moments about 

500X1-10 — (?X 10 = 0; 



84 Strength of Material. 

and Q = 333^, the other supporting force being 500 — 333-^; 
or P = 166f . But we can get P in another way, for 

and 

B. M. = ^Fdx = j (P — 5a;2) dx =^ Px — ^-\- 0^, 

o 

C^ is eqnal to zero, for the bending moment at the origin is 
zero, and M is also zero at the right end of the beam, so that 
if we snbstitnte a; = 10 and solve the equation 

p ^ 10 _. 1X1000^0 

we get P = 166f as before. Putting this value of P in the 
equation for S. F. and B. M. we get 

S.F. = 166| — 00^2 (1) 

and 

B.M. = 166fa; — ^, (2) 

which equations give the curves ACB for shearing (Fig. 35) 
and ODX for bending. 

If we put (1) equal to zero and solve for x we get the point 
on the beam where the shearing force is zero; and if we put 
this value of a; in (2) we will get the maximum B. M., for 
the maximum B. M. occurs where the S. F. is zero (see Art. 
28), for by the principles of maxima and minima, the first 
derivatives of the B. M. with respect to x, put equal to zero 
will give us the shearing equation (1) equal to zero 

diBM) ^ leet - 5.- = ; from which .• = J^^ = ^ 
dx ^ ' V3X5 VS 

the value of x where B. M. is a maximum. 



Strexgth of Material. 85 

Examples: 

1. A beam 10 ft. long, supported at the ends, carries a load 
of 1000 lbs. 4 ft. from the left end. Find curves of S. F. 
and B. M. 

2. A beam 5 ft. long, fixed at the left end and unsupported 
at the right end, carries 1000 lbs. at the right end. Find 
curves of S. F. and B. M. 

3. A beam 10 ft. long, supported at the ends, carries a uni- 
form load of 100 lbs. per ft.-run. Find curves of S. F. and 
B. M. ; give values at ends and center. 

Ans. S. F., ends =b 500; center 0. B. M., ends 0; center 
1250 Ib.-ft. 

4. A beam 5 ft. long, fixed at the left end, unsupported at 
the right end, carries a uniform load of 100 lbs. per ft.-run. 
Find curves of S. F. and B. M. ; and give values at ends. 

Ans. S. F., left end 500 lbs. ; right end 0. B. M., left end 
—1250 Ib.-ft.; right end 0. 

5. A beam 10 ft. long, supported at the ends, carries 400 
lbs. 4 ft. from the left end, and 600 lbs. 6 ft. from the left 
end. Find curves of S. F. and B. M., and give values at the 
ends and center. 

Ans. S. F., left end 480 lbs.; right end 520; center 80. 
B. M., left end lbs.; right end 0; center 2000 Ib.-ft. 

6. What is the longest steel bar of cross-section of 1 sq. in. 
that can be supported at its center without being permanently 
bent, the greatest allowable bending moment for the bar being 
2000 Ib.-ins. ? 

Ans. 19|i ft. 

7. A beam 20 ft. long, supported at the ends, carries 2000 
lbs. 5 ft. from the left end, and 5000 lbs. 4 ft. from the right 
end. Find curves of S. F. and B. M., and give values at the 
ends and at the loads. 

Ans. S. F., left end 2500 lbs.; right end 4500 lbs.; be- 



86 Strength of Material. 

tween loads 500 lbs. B. M., end ; first load 12,500 Ib.-ft. ; 
second load 18,000 Ib.-ft. 

8. If the beam of example 7 were loaded with. 200 lbs. per 
ft.-run, find the curves of S. F. and B. M. and give values at 
ends and center. 

Ans. S. F., ends ± 2000 lbs.; center 0. B. M., ends 0; 
center 10,000 Ib.-ft. 

9. A round steel rod of 2 ins. diameter can only withstand 
a bending moment of 6 ton-ins. What is the greatest length 
of such a rod which will just carr}^ its own weight when sup- 
ported at the ends? 

Ans. 29.2 ft. 

10. A beam 20 ft. long, supported at the ends, carries a 
uniformly distributed load of 5 tons, and a concentrated load 
of 5 tons, 4 ft. from the left end. Find curves of S. F. and 
B. M. and give values at the center. Where is the greatest 
bending moment and give its value? 

Ans. S. F., center — 1 ton. B. M., center 22J ton-ft. 
B. M., maximum 24^ ton-ft., 6 ft. from left end. 

11. A pine beam 20 ft. long and 1 ft. square floats in sea 
water. It is loaded at the center with a weight just sufficient 
to immerse it wholly. Find curves of S. F. and B. M., and 
give maximum values. A cu. ft. of pine weighs 39 lbs., of 
sea water 64 lbs. 

Ans. Maximum S. F., 250 lbs. Maximum B. M., 1250 
Ib.-ft. 

12. A beam 54 ft. long, supported at the ends, is loaded 
with 15 cwt. per ft.-run, for a distance of 36 ft. from the left 
end. Find the curves of S. F. and B. M. What is the maxi- 
mum B. M. and the B. M. at 6, 12, and 36 ft. from the left 
end? 

Ans. B. M. maximum = 216 ton-ft. ; B. M.^ = 94.5 ton- 
ft. ; B. M.12 = 162 ton-ft. ; B. M.g^ = 162 ton-ft. 



Strength of Material. 87 

13. A steel beam 5 ft. long is fixed at one end^ unsup- 
ported at the other end, and is of rectangular section 2 ins. 
wide and 3 ins. deep. What weight at the free end will de- 
stroy the beam if the limiting stress is 24 tons per sq. in. 

Ans. 1.2 tons. 

14. The buoyancy of a floating object is at the ends, and 
increases uniformly to the center, while the weight is at 
the center and increases uniformly to the ends. Find the 
curves of S. F. and B. M. and give maximum values in terms 
of the displacement, D, and the length, /, of the object. 

Ans. S. F. maximum := — ^ ; B. M. maximum = — ~~ 



88 Strength op Mateeial. 



CHAPTER YIII. 

Slope and Deflection. 

43. The slope at any section of a loaded beam is the angle 
between the tangent to the neutral line at that section and the 
straight line with which the neutral line would coincide if the 
beam were not bent; or, slope is the angle between our axis 
of X and the tangent at any section to the curve into which 
the beam is bent. 

The deflection at any section of a loaded beam is the dis- 
tance from the axis of X to the point where the neutral line 
pierces that section; or^ it is the ordinate at that section of 
the neutral line (see Fig. 36). 

From Chapter lY we have the general formula for bending 

-y = -p- , irom wnicn K = -^ . 

E being a constant and I also, in this case, as we are consid- 
ering beams of uniform cross-section; M, of course, varies at 
different sections of the beam. 

By calculus the formula for the radius of curvature at any 
point of a curve given by its rectangular equation is 



R 



[> +(!)■] 



d'y 
dx' 



Equating these two values of R we have 

— -t±(i2, 



Strexgth of Material. 89 

which by integration will give ns the equation to the curve 
into which the beam is bent. As this integration would be 
somewhat complicated^ and as in properly built structures the 
dimensions of the different pieces of material are such that 
the bending is very slight, we can without appreciable error 

simplify the operation considerably as follows : J- being the 

tangent of the angle that the bent beam at any point makes 
with the axis of X, is a very small fraction, and being 
squared in equation (1) becomes so small that it may be neg- 
lected, in comparison with unity, so that equation (1) for all 
practical purposes becomes 

EI 1 ^,r d'y 

dx' 
and integrating this equation we get 

EI^ = SMx, (2) 

which will give us the tangent of the slope at any point when 
we substitute for M its value in terms of x as found in Chap- 
ter YII, and integrate. The integral, ^Mdx is called the 
slope function, and we will designate it by ;S'. When S is 
divided by EI we have the tangent of the angle of slope. 
If we write equation (2) 

ax 

and integrate again we get 

EIij=^Sdx. (3) 

This is the equation of the curve into which the beam is bent 
and the value of y given by this equation is the ordinate of 
the neutral line at any section distant x from the origin, y 
is usually negative. 



90 



Steength of Material. 



44. In integrating to get values for equations (2) and (3) 
we must not forget the constants of integration. 

If a beam is fixed at the origin^ the slope there is zero, but 
a beam supported at the ends and loaded will bend into a 
curve like that of Fig. 36. Clearly the slope is greatest at 
the ends and there will be a constant of integration for equa- 
tion (2). We will usually know from the way in which the 
beam is loaded a value of x, at which the slope is zero. 

For example, with a uniform load the beam of Fig. 36 
would bend so that at its middle the slope would be zero, 
therefore, if we substitute (having integrated the right mem- 



k~-jc — i 



H 




Fig. 36. 



ber) for x, in equation (2), half the length of the beam and 
put the equation equal to zero, we can solve for C , the con- 
stant. If we do not know a value of x where the slope is 
zero, we will know a point where the de-flection is zero (one 
of the supports for example), and substituting in equation 
(3,) the value of x for this point, solve for the constant of 
integration of equation (2). An examj^le will be solved illus- 
trating the method. 

There is usually no difficulty in finding the constant of 
integration for equation (3), because the deflection of a beam 
at the origin, whether the beam be fixed or supported there 
and no matter how loaded, is zero. Of course a beam could 
be propped up somewhere along its length, so that the left 
end would not rest on its support. 



Strexgth of Material. 91 

45. A beam, 10 ft. long^ supported at the ends, is loaded 
with a -aniform load of 25 lbs. per ft.-run. Find the slope 
and deflection. 

Here the bending is perfectly symmetrical. Let Fig. 36 
represent the bent beam, being the origin and OX the posi- 
tion of the neutral line before the load was applied, the dotted 
line representing its position after the application of the load. 
P = Q = 125. 

L = — 2b, 

F = — 2dx 4- C = — 25.T + 125. [C = P], 

M = -^ + 12ox+[C = 0], 

^^^ = -¥ + ^+^.- a) 

Knowing the bending is symmetrical, -^ = where a; = 5 
(the middle), so 
,^_25xi3o_^m^5_^^^ ..<7, = -1041|. 

In case we do not know the bending is symmetrical, we carry 
Ci down through the next integration for deflection which 
gives 

^^^ = -~2r+~"6 VCi^^VC,= (). (2) 

Co is zero for the deflection is zero at the origin. The deflec- 
tion is also zero where x = 10 (the other support), and sub- 
stituting this value we get 

25 X 10000 125 X 100 , ^.^ 
24 "^ 6 ~r-^^^i, 

from which C^ = — 1041f as before, and equation (1) be- 
comes 

T^j- dy 2ba? , 125.^' ,^,,., 



92 Strength of Material. 

from which by substituting the abscissa of any section of the 
beam for x and dividing the result by EI we get the tangent 
of the angle of slope for that section. Substituting the value 
of Ci , in equation ( 2 ) gives 

EIy = ~-^A 6 104Ha;, (4) 

from which the ordinate of the neutral line (curve of bend- 
ing) at any section may be found. In finding the value of 
this ordinate^ or of the slope^ care must be taken to use the 
same units throughout ; for example, if we have a steel beam 
and E is given as 30,000,000, it is in lbs. per sq. in. ; we must 
therefore use x in ins., find / for the section in in. units, and 
reduce the bending moment to in.-lbs. ; this will give us the 
deflection in ins. 

It is convenient to solve this problem by using letters to 
represent the numerical data and to make the substitutions 
later in the equation which gives the desired result; for ex- 
ample, in the above problem let the load per ft. -run equal 
" w " and the length of the beam " a," then 

F — — lox -\- C = — iux-\- F = — tux-\-^ , 
wax 



/ 
dx 



„^ dy tux"^ wax' .^ 



Kt J , ^^ (f J 



o = --^+-i^+^.. .-.^^^-l- 



^-r wx" , wax-" wa'x , ^W ^^ 

By this latter method, the particular result desired may be 
obtained without doing the numerical work required for the 



Strexgth of Material. 



93 



other equations. In the above beam the maximum slope is 
at the ends, and the tangent of the angle is — . 



lua" 



UEF The 

maximum deflection occurs at the middle of the beam, and 

there?/ — —-^^^ 

46. A beam, supported at the ends, carries a single concen- 
trated load TT at a distance " a '' from the left end and " b " 
from the right end. Find equation for slope and deflection. 

In this problem the beam will be bent as shown in Fig. 37, 
and the elastic curve will consist of two branches, the part 




Fig. 37. 

from to the weight and that from the weight to the other 
end of the beam. The shearing force on each part will be 
constant, but the values will be different and they will have 
different signs. We must consider the two parts separately. 
With the origin at 0, the bending moment for any section 
EH to the left of the load W is 
Whx 






a-\-h ' 



+ 0,, 



dx 2{a^b) 

For the part to the right of the load W we will take our 
origin at the right end, then, letting x-^ be the distance from 



94 Strength of Material. 

the right end to any section H'H', the bending moment will be 

the sign being negative because the moment tends to turn the 
right end of the beam counter clockwise, or in the opposite 
direction to that of the other end; we have then for the riglit 
end 

Wax, 



M= — 



a-Yl ' 



cly _ Wax,' 

^^ dx -~ 2(a + b) + ^ 1 • 

If we move the origin for these equations back to the left end 
of the beam, b}^ substituting \x — (^ + &) [- for .t^ , we get 

Wa{a-^h-x) 
a^l ' 



J.J dy_ _ Wa{a -{-i — xj , 

^^ dx -~~ 2C^ + ^) +^1' 

Now for the section under W the slope and deflection is the 
same using either the equations for the part of the curve to 
the left of W, or those for the part to the right of W, so we 
can put the corresponding values equal to each other when 
x=i a, 



2(a-\-l)) ^"1- 2{a + h) 



OT,C,-C\=^ . (1) 



WM' _ Wa¥ ^, ^ 



6ia + b) -r-i-— 6(a + 5) 



or,C.,+ 0'.i=^^^i^. (2) 



Strength of Material. 



95 



From the simultaneous equations (1) and (2) we get 



C, = - 



and substituting these values of C^ and C\ we get 



dy 



El ^ - 



for the left, end, 
Whx'^ Wahia- 



2&) 



dx 2(a + &) 



6(a + 6) 



EIy = 



Whx^ Wah{a + 2h)x 



6(a + &) 



6(a + &) 



EI ~ 
dx 



for the right end, 

_ Wa{a + & — x)2 
2(a + h) 



Ely = 





+ 


Wa&(2a + &) 
6(a + 6) 


Wa{a + 6 - 


a;)3 




6(a + 6) 






Wa&(2a 


+ 6) (a + 6 - cc) 



6(a + h) 

and from either set of these equations we get the deflection 

at the load to be 

_ WaW 

y~ 3EI{a + l))' 
Since the load is not at the middle of the beam, the maximum 
deflection will occur in the longer segment, and at the section 
of maximum deflection the slope will be zero; therefore, sup- 
posing " a '' to be greater than " b," if we put the equation of 
the slope for the part of the curve to the left of W equal to 
zero and solve for x we will get the abscissa of the section of 
maximum deflection; and substituting this value of x in the 
equation for the deflection will give us the maximum deflec- 
tion, as follows: 

Whx^ Wa'b(a + 2h) . 

6(a+&) ^'''' 



— 



2{a+h) 
and this value of x substituted in 

Whx^ Wad (a 



,^^aia±U) 



EIy = 



2'b)x 



6(fl + &) 6{a + h) 

shows the maximum deflection to be 



ymax — 

or, if a = &^ 



Wh 



3EI(a + l)) 



'a(a+2h)^ 



ymax — 



Wa' 

6 Br 



96 Strength of Material. 

Examples: 

1. A steel beam, 10 ft. long, supports a concentrated load 
of 25 tons at its middle. What is the deflection under the 
load if / in in. units is 84.9 and E = 30,000,000 ? Find 
equation of elastic curve. 

Ans. .077 — ins. 

2. A steel beam of I section, 20 ft. long, 8 ins. deep, 3 ins. 
wide, with flanges and web each -J in. thick, is used to sup- 
port a floor weighing 200 lbs. per sq. ft. The beams are sup- 
ported at the ends and spaced 3 ft. apart. What is the maxi- 
mum deflection ? E = 30,000,000. Find equation of elastic 
curve. 

Ans. 1.3 — ins. 

3. A beam, supported at the ends, carries a uniform load of 
150 lbs. per ft.-run. It is 10 ft. long, 6 ins. deep, and 4 ins. 
wide. E = 1,200,000. Find equation of elastic curve and 
the maximum deflection. 

Ans. .51 — ins. 

4. A beam 12 ft. long, 8 in. deep, and 12 in. wide, is sup- 
ported at the ends and carries a load of 1200 lbs., 3 ft. from 
the left end. E — 1,200,000. Find the elastic curve and the 
maximum deflection. 

Ans. .12 + ins. 

5. A steel beam, 60 ft. long and weighing 100 lbs. per ft., 
carries a concentrated load of 14,400 lbs. at the middle. If 
/ in in. units is 1160, find the maximum deflection. E = 
30,000,000. 

Ans. 6.38 — ins. 

6. A beam, 20 ft. long, supports a floor weighing 100 lbs. 
per sq. ft. The beams are spaced 4 -ft. apart. Supposing 
the section of the beam to be square, find the side of the square 
in order that the deflection may not exceed ^ in. ^ = 700 
in.-tons. 

Ans. Side is 12.18 ins. 



Strength of Material. 97 



CHAPTEE IX. 
Slope and Deflection" — Continued. 

47. Example : A beam of length " a " is fixed at the left 
end and unsupported at the right. It is loaded with a uni- 
form load of w lbs. per ft.-run. Find the equation of the 
elastic curve. 

L = — w, 

F= — wx-\- C = — lux^wa (P = whole load) , 

M= -'^ + wax+ \_G,=-'^{C, is found in Art. 40), 

(beam fixed at origin) , 

(beam fixed at origin) . 
The maximum deflection obviously occurs where x =^ a. 

If we suppose the free end of the above beam is propped 
up to the same level as the fixed end. The prop now sup- 
ports some of the load so we do not know P. 
L:= — w, 
P = — wx-{- C = — wx-\- P, 

M = -f\px+C,. 

We know the bending moment at the end supported by the 
prop is zero, so putting If = and substituting x = a 

from which „ 



Strength of Material. 



Substituting this value of C^ we get 






Pa, 



EI^ = -^^-\-^^^^-Pax+[C.= Q 



dx— 6 ^ 2 ^ 2 
„^ wx^ , Px^ , wa^x^ Pax^ 



(beam fixed at origin) , 



2 + [C^3 = 

(beam fixed at origin) , 




Fig. 38. 



We can now Find P, for the deflection is zero where x ^ a, 
and all the other constants are determined. 

iua^ , Po? , 

from which 

htm 



wo!" Pa^ 



P = 



and substituting this value in the equations will give us the 
curves desired and the slope and deflection. The curves are 
shown plotted in Fig. 38. AB being the curve of S. F., and 
CT)X that of bending moment. The B. M. at the origin is 



negative and equal to 



7oa" 
"8" 



. The greatest positive B. M. 



Strength of Material. 99 



occurs where the shearing force is zero at x =^ -^ , and substi- 

o 

tuting this value in the equation for B. M. gives the greatest 

Q 7/1/7 2 

positive B. M. equal to . The maximum B. M. is there- 

fore the negative one, and the beam will break if overloaded, 
at the fixed end. 

It will be noticed that the B. M. is zero at the free end, 
and also at a point between there and the fixed end. Putting 
the equation for B. M. equal to zero and solving for x, gives 

x = a and ^^^ = ? • The point where x ^^ is called a virtual 

joint, lecause, if a beam loaded and supported in this way 
had a hinge at this point, the bending moment would not 

cause it to turn. Putting -^ = 0, and solving for x gives 

X =: .58a and a value greater than the length of the beam. 
Where x = .oSa the slope is zero, and this is the point of 
maximum deflection. Substituting this value of x in the 
equation for deflection gives 

_ MiSwa" 

ymax — ^j . 

Compare the curve of B. M. for this beam with that in 
Fig. 33. As soon as the prop under the free end of this beam 
begins to bear a part of the load there exists a positive B. M. 
and the virtual joint which is then near the right end of the 
beam, moves toward the fixed end as the right end is propped 
up and the load on the prop increases; but the results due to 
change in conditions are readily followed. 

48. A beam " a " ft. long is fixed at both ends and carries 

a concentrated load W at its middle. Here P = ^ = -— . 

z 

W 

.'. S. F. = ^r- and is constant from the ends to the load, but 
z 

has opposite directions on opposite sides of the load; there- 



100 



Stejbngth of Material. 



fore, as in Art. 44, we must consider the two parts of the 
beam separatel)'. To the left of the load, then, 

Wx 

M= ^ [- C , ((7 is not known), 

EI^^=^+Cx-^[C, = (beam fixed at end) . 
The loading being symmetrical, the slope at the middle is 
zero, substituting x =: -^, 



= 4^+ 0(1 



or, C 



Wa 

8 




The equation of the curve of bending to the left of the load, 
then, is 

,^ Wx Wa 

M=~^ W 

This is an equation of the first degree, and therefore repre- 
sents a straight line {EF in the fig-ure). We know the B. M. 
is symmetrical, and that the curve for the right end will have 
equal ordinates, but of opposite sign (the curve being FG in 
the fig-ure). j^ow it is clear we cannot embrace both of these 
straight lines in one equation of the first degree, and for this 
reason we must consider the two parts separately. Before 



Steje^^gth of Material. 101 

proceeding, we will notice that the bending moment has the 
same value with different signs at the ends and in the middle. 
This shows that the beam is as likely to break at the ends as 
in the middle if overloaded, and also that the virtual joints 
are at quarter span. 

,, Wx Wa 

M^^ T' 

Wx^ Wax' , ,^ 
EIy = -^ j^ + [^, = 0. 

The maximum deflection is at the middle and equal to 

_ W 

49. The Cantilever Bridge. — In the cantilever bridge the 
joints are placed at the points where the bending moment 
would be zero if the bridge were continuous over the whole 
span. We will consider a beam fixed at hoth ends, having 
two fixed joints and carrying a uniform load of w lbs. per 
ft. -run. There being joints at F and G (Mg. 40) we must 
consider the part FG as a separate beam supported at the 
ends, the load on it being equally divided and supported at 
F and G by the beams OF and GX. Erom the conditions 
we could assume P and Q ivith this loading to be equal to 

-^ — ' "*" ^ , but we can get their values in another way : 

L = — w, 

F = — wx -\- C ^ — wx-{- P, 

The bending moment at the joint is zero, so 

= -'^ + Pa + C,, (1) 

= --^^+^^+P(« + 6) + C,, (2) 



10^ 



STREi^GTH OF MATERIAL. 



and from equations (1) and (2) we get 



substituting. 



(2a + &,). 



and Ci = 



^oa 



(a+h). 



w 



wx + ^{2a + h), 



M = -'J^+^iU+l)x 



iva 



{a+l). 



The joints being where B. M. is zero, the curves of S. F. and 
B. M. will be continuous as shown in Fig. 40, but for slope 
and deflection we must consider each part as a separate beam, 




the parts at ends having in addition to their uniform load a 
concentrated load at their ends, F and G, equal to half the 
total load on the middle part. For equal strength, the bend- 
ing moments at the ends and middle of the above beam should 
have equal values and this requires " a '^ to be equal to " c," 

and ,„ 

wa , , , ^ w¥ 

whence 

46t24-4&fl=&2; or, 4a2+4&a+&2=2&2; or, {^a+lY—^V', 
or the square of the whole span must equal 2&^, and the joint 



must be distant from the middle of the beam 



up an 



Strexgth of Material. 



103 



50. Travelling Loads. — If a load W moves from left to 
right across a beam of length " a/^ the positive shearing force 
for any position of W is equal to P (Fig. 41). P is ob- 
vioush' greatest when W is just over it, and as W moves across 
the beam. P"s value drops nniformly until when W is just 
over Q^ P's value is zero. We can then represent the change 
of the positive S. F. by the line AX, and by the same reason- 
ing the change in the negative S. F. would be represented 
by OB. Considering the B. M. we know that for any posi- 
tion of W the B. M. is greatest for the section under W and 




-P. ^(o-x-j 



Fig. 41. 



for that section is i¥ = Pa; =— (a — x) x. By the principles 

of maxima and minima, we can find the value of x for which 
M is a maximum by putting the first derivative of M with re- 
spect to X equal to zero and solving for x, as 

cm w 

dx a 
which gives 



= -^ {a — '2x) =0, 



_ a 

the value of x, or the position of W which gives the maximum 
B. M. If the travellino- load is continuous, such as a train 



104 



Strength of Material. 



of cars crossing a bridge, the maximum value occurs when, 
if the train is long enough, it extends completely over the 
bridge; if it is not long enough for this the maximum positive 
S. F. occurs when the whole train has just got on to the bridge 
and the maximum B. M. when the middle of the train is just 
over the middle of the bridge. 

51. Oblique Loading. — Loading is said to be oblique when, 
for any cross-section of a beam, the plane of the external 
bending moment does not pass through a principal axis of 
the section. 




Let Fig. 42 represent the section of a beam and let the 
arrow-head marked M represent the direction of the B. M. 
The axis of X is perpendicular to the plane of the paper 
through 0, and the axes of T and Z are shown. Eesolve M 
into components parallel to the axes of Z and Z. The direc- 
tion of the fiber stress due to bending is perpendicular to the 
plane of the paper, and that part of it,, due to M sin a, which 
acts normal to the side AD of the section is 

M sin a . 2 
P^ = 7 > 



Steexgth of Material. 105 

which is obtained by substituting M sin a, the component mo- 

ment, in t_ _ r^ ^ the general equation for bending, ly being 

the moment of inertia of the section about the axis of Y, 
and z the distance of the line AD from the axis of Y. The 
normal fiber stress perpendicular to the plane of the paper 
on AB is 

M cos a 

Py = — T — • y- 

Applying the principle of superposition the total fiber stress 

at A would be 

M sin a . 2; , M co&a .y 
,P = Py + P^ = 7 \ 7 . 

52. The worh done in bending a beam is evidently equal to 
Md, where M is a uniform bending moment and 6 is the angle 
of slope at the ends. If I is the length of the beam and R the 

radius of the curve into which it is bent, then 6 = ^-p f ^^^ 
from -J = ~ \^e have B = -jj= . Hence the work done by a 

uniform bending moment is equal to ^ etv • But bending 

moments are seldom uniform, so, for a variable bending 
moment 

^oi:\= —S^Pdx. 

Examples : 

1. A dam is supported by a row of uprights fixed at their 
base and their upper ends are held vertical by struts sloping 
at 45°. Water pressure varies as the depth, or L = — ivx. 
Find the equation for deflection. Length of upright " a/' 
What is the thrust on the struts ? 

Ans. Thrust = y horizontal pressure of water. 



106 Streitgth of Material. 

2. A railroad is inclined at 30° to the horizontal. The 
stringers are 10.5 ft. apart and the rails are 1 ft. inside the 
stringers. The ties are 8 in. deep and 6 in. wide. The load 
transmitted by each rail to one tie is 10 tons. What is the 
maximum normal stress in each tie ? 

Ans. 5744 Ib.-ins. 

3. A beam, 2a ft. long, fixed at the ends is uniformly loaded 
with w lbs. per ft.-run. Find the maximum deflection and 
the virtual joints. 

Ans. Maximum deflection := — 9aWT' ^i^^ual joints 
where ic = a(l it Vi)- 

4. A beam, " a " ft. long and fixed at the ends, carries a 
uniformly increasing load from zero at the left end to w lbs. 
per ft.-run at the right. Find the maximum deflection and 
the virtual joints. 

Ans. Maximum deflection := — 010-7^7- • Virtual joints 

, a ^ 2a 

where ^' = -^ ^i^"- ""5 • 

5. A beam, " a " ft. long, fixed at one end and the other 
end propped up to the same level, carries a uniformly decreas- 
ing load from w lbs. per ft.-run at the fixed end, to zero at 
the propped end. Find the equation for deflection, the posi- 
tion of the maximum positive B. M., and the position of the 
virtual joints. 

Ans. Maximum positive B. M. where x = aVi- Vir- 
tual joints where x ^ aVf . 

6. A beam, " a '' ft. long, fixed at both ends, carries a single 
concentrated load TF at a distance "-d" from the left end. 
Find the deflection under the load. 

Ans. b- g^j^3 



Strength of Material. 107 

7. If the load of the beam in example 6 had been at the 
middle of the beam, what would have been the maximum 
deflection ? 

8. A single load of 50 tons crosses a bridge of 100-ft. span. 
Draw the curves of maximum S. F. and B. M., and give the 
values of those quantities at half and quarter span. 

9. A train, weighing 1 ton per ft. -run and 112 ft. long, 
crosses a bridge of 100-ft. span. Draw curves of maximum 
S. F. and B. M., and give values at half and quarter span. 

10. A steel shaft carries a load equal to h times its own 
weight, first uniformly distributed, second concentrated at its 
middle; considering it as a beam fixed at the ends, find the 
distance apart of the bearings that the ratio of deflection to 
span may be tto-q. 

Ans. (1) Span in ft. = 10.5 '/__^ 

^ ^ V k^i' 

(2) Span = 8.3 ^^r4^^ (d in ins.). 



108 



Strength of Material. 



i 



CHAPTEE X. 

Continuous Beams. 

53. A contin-aous beam is one which extends over several 
supports. In this chapter we will consider as heretofore only 
such beams as have a uniform cross-section and, as in all 
practical constructions the supports are adjusted so as to 
allow as little strain as possible in the material, we will 
assume all the supports to be at the same level. It is obvious 




that the bending moments at the intermediate supports will 
be negative as at these points the conditions are- similar to 
those of a beam balanced over a single support. If the end 
spans are short we may have the supporting forces at the 
ends equal to zero, and if the ends are " anchored'^ (fastened 
down to the support) we may have a negative supporting 
force, that is one acting in the same direction as the loads. 
In Fig. 43, we see approximately the -curves of bending mo- 
ments and shearing force for a continuous beam carrying a 
uniform load and having five equally spaced supports. There 
are two virtual joints between supports except for the end 



Strength of Material. 109 

spans, and we will find that the value of the deflection in 
each span is somewhere between that for a uniformly loaded 
beam of span length supported at the ends, and one nniformh' 
loaded of span length fixed at the ends. The difficulty in 
working with continuous beams lies in finding the supporting 
forces, for the usual method of taking moments will not do, 
there being two unknown quantities in each equation, and the 
equations reducing to identities when we attempt to solve 
them. For beams having only three supporting forces, the 
solutions are not difficult, as will be shown ; but as the number 
of supports increases the calculations become more and more 
tedious. 

54. A beam of length 2a, carries a uniform load and is 
supported by three equidistant supports. A beam of length 
%a carrying a uniform load of iv per unit length and sup- 
ported at the ends, has a maximum defiection equal to 

^IWf— (see Art. 45). A like beam supported at the ends 

and carrying a concentrated load R at its middle has a maxi- 

mum deflection equal to — l^^FJ ^^'^^- ^^)' -^^^ ^^^ 

above continuous beam may be considered as a uniformly 
loaded beam which has a prop at its middle by which the 
middle point is propped up to the same level as the end sup- 
ports. Obviously then the deflection at any point will be 
that due to the uniform load minus that due to the thrust 
of the prop, and as the prop deflects upward we have 

R{2ay _ ^w{2aY 
^— 4.8EI 3MEI ' 

or R, the thrust of the prop, or the middle supporting force is 

-R = I wa. 

7 



110 



Strength of Material. 



The supporting forces at the ends are obvionsly equal, so 



1 



P = = 



2 



^ ^wa. 



Having the supporting forces we have now no difficulty in 
getting the equations for slope and deflection, for taking the 
origin at the middle; the B. M. for any section distant x from 
the origin is, by definition (remembering that the part of R 

7?\ 

which is due to the loading on one side of the origin is -^ j. 



rr:^-0^ 






-; 



cc 



-^:- 



T- ^w<^. 



.7-r\. 



a. -- 



Fig. 44. 



T" 
i 



Q-^ 



W€L. 



M^EI 



dy Rx 






R X , ^r 

-^ X — wx .-^ 4- Mq 



wx"" 



^/2/ = ^ -H-* + i/.^+ [C, = . . . . I 



f from symmetr}^ 
the slope is zero 
at the origin. 



deflection is zero 
at the oriffin. 



Substituting the value of R we have for the B. 31. at any 
section 

1DX? 

M = ^wax — -^ + i¥o . 
And as the B. M. is zero at the ends, we have, substituting 
a; = a, the value of Mq equal to — -q- : or 



r 

"I 

i 

i 



H 



M = %ivax 



wx 
2~ 



10 a' 



* Strexgth of Material. Ill 

putting this value of M equal to zero and solving for x gives 
us a; = zb -^ as the position of the virtual joints. Integrating 
we get the equations for slope and deflection 



and 









To show how carefully the level of the supports must be ad- 
justed, suppose there is left a deflection over the middle sup- 
port, equal, let us say, to J of the total deflection which would 
occur were there no middle support; then 

^ ' 3S4.EI ~ 3S4.EI 4:8EI ' 

from which U — ^ ^^^^ , or one-half the total load. If this 

deflection were in the other direction, or — \ we would have 
JK = f the total load. Remembering that the deflection in 
any case is very small when we see that a variation of it \ 
up or down causes U to change from f to -J the total load, 
we can understand how carefully the supports must be ad- 
justed to the same level. 

55. With concentrated loads we must proceed in a different 
way. Suppose a beam which is supported at the middle and 
ends carries a concentrated load "PF midway between the sup- 
ports, as represented in Fig. 45. Taking the origin at the 
middle as before and considering the right half of the beam, 
the bending moment at any section KIL between the origin 
and the load is (Art. 46), 

i/-^/g=-If(|--a;) + e(«-^), (1) 



11^ 



Strength of Material. 



Integrating, 



EI 



dx 



Wax , Wx" , ,, Qx' , ^^ , 



and 



^^/, = _i^ + ^V^-|^ + [C,==0. (8) 



Por a section between W and the end of the beam 



-w 



M=EI^, = Q{a-x). 



1 W 



(4) 



f 



CL 



VR-'h 






Integrating, 



Fig. 45. 



^^l = «--¥+[^= 



a=^w 



(5a) 



The slope under the load is the same in both branches of the 
curve, therefore, substituting a; = -^ in equations (2) and 
(5a), and equating them we have 
Wa^ . Wo" , Qa" Qa" Qa" Qd 



4+8+2 
Substituting, 



2 



Integrating, 



+ C, or (7= 

dy _ Q^ W 

dx-^^''~~ 2 ~- 8 • 

Qax" Qx' Wa'x 



m = ^-\ 



+ 10, 



8 • 

(5) 

(6a) 



ii 



Strength of Material. 113 

The deflection under the load is the same in both branches, so 

as before, using equations (3) and (6a) we have 

_Wa^ Wa^ Qa^ Qa' Qa^ Qa^ ^^s 

16-^-48+8 ~"18 -^— 48 ~-T6 + ^i' 
or, 

^ _ Wa' 
Substituting, 

i^/y--^- — _^_+___. (6) 

-Now the deflection at the right support is equal to zero, and 
putting equation (6) equal to zero, substituting x = a, solv- 
ing for Q, gives Q = ^F. From symmetr3r, P equals Q 
and R must support the rest of the load, hence 

P = 2F-(P + 5)=J^i-F. 
The methods of this and the preceding article will cover 
most of the cases of continuous beams found in actual prac- 
tice, but there are many cases where beams have more than 
three supports, and for these a method known as the 
"Method of Three Moments- must be employed This 
method came into use about 1857, and is a general method for 
obtaining the supporting forces. 

56. Theorem of Three Moments—Let Fig. 46 represent 
the part of a beam over any three consecutive supports, and 
let the loading be uniform between adjacent supports. Take 
the origin at P and let the bending moments at P, R and Q 
be M^, M, and M, respectively, the distance between sup- 
ports to be a and a^ , the loading between P and P be ?^ lbs 
per unit length, and between R and Q, w, lbs. per unit length; 
then the bending moment at any section HE between P and 
22 IS (Art. 45), 



114 



Strength of Material. 



but C in this case is equal to M^, so 



wx' 



Integrating, 



%_ 



El ± = M,x -V P ~, 



X wx 



dx 



F + [6' = ?, 



(1) 

(3) 



and 



Ely = M,^^P^-'^+Cx^iC, = 



J>4 



(level supports), (3) 



^1 



K---X- 



■^- 



t 



7? 

Fig. 46. 



Q 



Now ?/ = when x=^ a, hence 



= 



i/,^^ , P«3 Z£/«* 



+ 



-^+0a. 



Substituting, 



and 



^^1=^.^+*-^' 



ij/^rc' P:»^ ?z;a;* 



.6- ^ 



2 "~ 6 + 



M^ax Pa^x 



Pa^ . wa^ 



+ 



24 



24 • 
(5) 



^^2/ - 2 + 6 ~ 24 ^ 2 ~ 6 ^24 
Prom (1), if we let x =^ a, we get M2 in terms of M^ , or, 

ilf, = 21f, + Pffl-if^. (6) 



Now if we imagine the beam reversed and use Q for the 
origin, we can in the same way find the equation for bending 
moment, slope and deflection; for the part of the beam be- 



for the left branch, 






for the right branch, 


M.^Pa- 


2 - 




M.„ 




= M,-^-Qa 


w,a^ 

2 ' 


i%a Pa' 

2 + 3 " 


8 - 


slop 


e over 


R 


_ M,a, 


Qa,' w,a,^ 


- 2 


3 •" 8 



Strexgth of Material. 115 

tween Q and R, and over R the values of the bending moment, 
slope and deflection will be the same for both branches of 
the curve. We can therefore equate these values if we 
substitute a; = a in those of the left branch and x = cfi in 
those of the right. Making these substitutions and equating 
we sret 



(7) 

.(8) 

If we substitute the values of Q and P from equations (7) 
in equation (8) and reduce we get 

M^a + 2Jf 2 {a + a^) + M,a^ = — ^^^' + ^A' . (9) 

This is known as the equation of the three moments, and by 
using different sets of three consecutive supports we can get 
as many equations, less two, as there are supports. From our 
knowledge of conditions at the ends of the continuous beam 
we can get two more equations involving the moments over 
the supports, and thus having as many equations as there are 
unknown quantities we may find all the supporting forces. 
For example, let the beam of Fig. 43 be supported at equal 
intervals and uniformly loaded with tv lbs. per unit length. 
From equation (9) 



or. 



M^a + 41f 2^ + M,a = — ^-^ 



M, + m, + M, = -^. (1) 



tea 



M, + m, + M, = --^. (3) 



116 Strength of Material. 

The bending moment is zero at the ends, hence ilf ^ = M 5 = 0. 
From (1) and (3) 

Substituting, we get from (2,) and (1) 

From (6), using the first three supports, 

M,^M, + Pa- 'if = - ^ . :.P = h\ »a. 

In the same way § =: |-| wa, and from symmetry we know 
/S' = and T = P, so the total load minus P+Q + T + 8 
= B from which R =^^^wa. 

Examples : 

1. A continuous beam of five equal spans is uniformly 
loaded with w lbs. per ft.-run. If the beam is 38 ft. long, 
what are the supporting forces and the bending moments at 
the supports? 

2. Find the side of a steel beam of square section to span 
four openings of 8 ft. each, the total load per span being 
44,000 lbs. and the greatest horizontal fiber stress not to 
exceed 15,000 lbs. per sq. in. 

3. Find the depth of a steel beam of rectangular section 
twice as deep as broad, to span three openings each 12 ft. 
wide, the total load on each span being 6000 lbs. and the 
greatest fiber stress allowed being 12,000 lbs. per sq. in. 

4. A continuous beam of three spans is loaded only on the 
middle span with a uniform load. What are the three sup- 
porting forces ? 

Ans. At ends — ^ . 
2u 



Strength of Material. 117 

5. A steel beam, / = 268.9 in in. units, is 36 ft. long and 
spans four openings, the end spans being each 8 ft., and the 
middle ones each 10 ft. Find the maximnm B. M. of the 
beam. What will be the uniform load per ft. to make a 
maximum fiber stress of 15,000 lbs. per sq. in. ? 

6. A continuous beam carries a uniform load. If the end 
spans are each 80 ft. what will be the length of the middle 
span in order that the B. M. at its middle may equal zero ? 

7. How much work is done on a beam 10 ft. long, 10 ins. 
deep and 8 ins. wide, which carries a uniform load of 250 lbs. 
per f t.-run ? 



118 



Strength of Material. 



GHAPTEH XI. 

Columns and Struts. 

57. When a prismatic piece of material, several times 
longer than its greatest breadth, is under compression, it is 
called a column or strut; a column being snch a piece placed 
vertically and carrying a static load, and all others being 
struts. 

A column or strut of material which is not homogeneous, 
or one on which the load does not act exactly in the geometric 
axis, will bend or buckle. For mathematical investigation 
we must assume our material homogeneous, and instead of 
assuming a slight deviation of the line of action of our load 
from the geometric axis, we will assume 
that the column has first been bent by a 
horizontal force, then such a load applied 
to its end as will just keep it in the bent 
form after the removal of the horizontal 
force. The assumptions we make will be 
three: (1), the column perfectly straight 
originally; (2), material perfectly homo- 
geneous; and (3), load applied exactly 
over the center of the ends. These condi- 
tions are never exactly fulfilled in prac- 
tice, so a rather large factor of safety must 
be applied. 

58. Euler's Formula for Long Columns. 

— We will choose a column with rounded 
or pivoted ends, so that they will be free 
to move slightly as the column is bent. 




Fig. 47. 



Strength of Material. 119 

Let Fig. -17 represent siicli a coliTmn, held in the bent position 
by the load W. Let the origin be at the center of the column, 
the axis of X vertical, and that of Y horizontal ; 8 is the maxi- 
mum deflection, and y the ordinate of the neutral line at any 
section HE distant x from the origin. From Art.. 43 we have 
the general equation of bending 

The bending moment for the section EH is, from the figure, 
M=W{S-y), 

SO 

h mcE 
grating, we get 



multiplying both members of this equation by 2 -^ and inte- 



dyV_2W3y Wf 

now y is zero at the origin and -^ is also zero there, the 

tangent to the curve of bending being parallel to the axis of 
X at that point, .'. the constant of integration being this 
tangent is zero. 

Extracting the square root of both members of equation 
(1) and, transposing, we get 

which, integrated, gives 

COS"' (^) = ^m ^ + 1^^ = 0- (2^ 

C2 is zero, for where a; = 0, y ^= and cos"^ 1 = 0. 



120 Strength of Material. 

Letting I equal the length of the column, when x—-~^ we 



which is E uteres formula for long columns with round ends. 
Transposing equation (2) we get for the equation of the 
elastic curve 



^ = .{l_cos(^{-^...)}. 







1/' 


f 


/: : 


/» 


L 1^^' 


c 


I: 




\ 




Fig. 48. 



59. A column loaded with W as per the 
above formula, will just retain any deflec- 
tion which may be given it, therefore this 
value of W is called the critical load, for 
the column will straighten out if there be 
any decrease in this load, and for any 
increase it will keep on bending until it 
breaks. This is the load then which puts 
the column in neutral equilibrium. The 
bending of columns is perfectly uniform, 
so we can derive from the formula of the 
preceding article, which is for a column 
with rounded or pivpted ends, the formula 
for columns with one or both ends fixed. 
Let Fig. 48 represent the elastic curve of a 
column with both ends fixed. There will 



have y = S, substituting these values in equation (2) we get 




cos-'O ^^^ . 3 . 




The angle whose cosine is zero is -^ . 




' ' 2 - \ EI ' 2 ' 


; 


from which we get 


i 


^="'f. 



Steexgth of Material. 121 

be two points of inflection, B and D, and as the bending 
is perfectly nniform these points will be at quarter the length 
of the column from the ends. The part BCD will represent 
the elastic curve of a column such as we considered in the 
preceding article, so the critical load of a column with fixed 
ends will be the same as for a column of half its length with 
pivoted ends, or. 

The part BCDE of Fig. 44 would approximatel}^ represent 
the elastic curve for a column with one end {E) fixed; so the 
critical load for it would be the same as for a column of two- 
thirds its length with pivoted ends, or 



W = 



4?^ 



This latter formula is not quite so accurate as the others, for 
the ends are not in the same vertical line. 

It has been shown by experiment that when the length of a 
pillar is greater than 100 diameters, the theoretical values are 
closely approached, while with shorter lengths these values 
are much too large. 

Columns having flat ends, if the ends are prevented from 
lateral movement, are considered as having fixed ends. 

60. As Eulers formula is applicable only in the case of 
very long columns, another formula has been obtained in dif- 
ferent ways by several different writers, which gives more 
accurate results for short columns. Obviously a column may 
be so short as to fail by crushing alone, in which case the 
crushing load would be W = fA where / is crushing strength 
of the material and A the area of the cross-section. For 
ordinary columns, then, W must lie between fA and the value 
given by Euler's formula. 



122 Strength of Material. 

The following formula is most used for short columns. 
If pi be the compressive stress due to W, and p^ the fiber 
stress due to bending, the maximum stress will be Pi + Pa 5 
this must not exceed the strength of the material, so for the 
limiting stress f z=z p^ -\- p^ . 

If A is the cross-sectional area of the column p^ = i^ . 

The equation of bending -^ = —^ gives us p2 = — ^ , remem- 
bering that y in this formula is the distance from the neutral 
plane to the extreme fiber of the section, 

' - A '^ I ' 

From Eig. 48, if the maximum deflection be 8, we have 

n 
M = W8, and as in symmetrical bending 8 varies as — we 

72 

have 8 = — (the constant is always a fraction so may be put 

in this way) ; substituting these values and remembering I is 
equal to A multiplied by a constant squared, or, I = AV, 
we have 

^ - A '^ cyi - A\^~^ ck'l' 
from which 

Af 



w= 



1+ ' 



ck' 



This is known as Rankin's formula for columns and struts 
with fixed ends. 



61. We see that when I approach-es zero, in the above 
rmula, W approaches Af, and as -j- inc 
proach the value given by Eulers formula. 



formula, W approaches Af, and as -j- increases, W will ap- 



Strength of Material. 



123 



The formula of Art. 60 is for a column or strut with fixed 
ends ; if both ends are rounded or pivoted^ we must divide the 
value of c by 4; if one end is rounded c is divided by 2, so 
that the formulae for different methods of end supports are 



Flat ends 



W- 



c¥ 



One round end W , 



Af 



1 + 



cJc' 



Both ends round W = 



^f 



4.P 
cJc' 



Values of the constants c and f^ for several materials are 



/ 

Hard steel 69,000 lbs. per sq. 

Structural steel 48,000 

Wrought iron 36,000 

Cast iron 80,000 

Wood 7,200 



m. 



c 

20,000 

30,000 

36,000 

6,400 

3,000 



Examples : 

1. A hollow cast-iron column, fixed at the ends, is 20 ft. 
high and has a mean diameter of 1 ft. It is to carry 100 
tons. Factor of safety 8. What is the thickness of the metal ? 

ilns. 1 in. 

2. What is the crushing load of a wrought-iron pillar 10 ft, 
high, 3 ins. in diameter, and with rounded ends ? 

Ans. 30 tons, nearly. 

3. What is the crushing load of a cast-iron column having 
flat ends, being 15 ft. long and 6 ins. in diameter ? 

Ans. 350 tons. 



124 Stbjjngth of Material. 

4. A wooden strut, 12 ft. long, supports a load of 15 tons. 
If the section' is square what must be the side of the square 
allowing a factor of safety of 10? 

Ans. 9.25 ins. 

5. A hollow wrought-iron column with flat ends is 20 ft. 
long, 10 ins. outside diameter, 7 ins. inside diameter. What 
load will it carry? 

Ans. 616 tons. 

6. A solid steel column with round ends is 6 ins. in diame- 
ter and 37 ft. long. What load will it bear? 

Ans. 90,000 + lbs. 

7. A square wooden column with fixed ends is 20 ft. long 
and carries a load of 9500 lbs., with a factor of safety of 10. 
What is the side of the square? (Eulers formula.) 

Ans. 3.24 ins. 

8. If in example 2 the pillar were of rectangular section of 
breadth, double the thickness, what sectional area would be 
required for equal strength? 

Ans. 9.4 sq. ins. 



Stkexgth of Material. 125 



CHAPTEE XII. 

Stress ox Members of Frames. 

62. We have thiis far considered tlie strength of single 
pieces of material; let ns now investigate the methods for 
finding the stress on any of the parts of a loaded structure 
consisting of several members. This can be done graphically 
or by calculation, and both ways will be considered. We will 
understand by the word structure an3-thing built of separate 
parts, called members, between which there is to be no motion. 
Each member must be strong enough to withstand all the 
forces to which it will be subjected without permanent de- 
formation. 

Forces are either external or internal. 

External forces acting on a structure are (1), the weights 
of the members; (2), the loads carried, such as the traffic 
crossing a bridge, rain and wind pressure on a roof, weight 
lifted by a crane, etc.; and (3), the supporting forces. The 
internal forces are the resistances offered by the members to 
distortion. 

The external forces acting on any member of a structure 
are its weight, the load and su^Dporting forces and the forces 
exerted on it by other members. 

We shall make use of the following rules for the equi- 
librium of a structure : 

(1) The algebraic sum of all the external forces taken 
together must equal zero. 

(2) All the forces, both external and internal, acting on 
any member must balance. 

(3) All the external forces on one side of any imaginary 



126 Strength of Material. 

complete section of a structure must balance all the internal 
forces on the same side of the section. 

(4) The algebraic sum of the moments about any axis of 
all the forces both external and internal on one side of any 
section through a structure must equal zero. 

All members of a structure are connected together by 
joints. 

By the word frame will be meant a structure which has 
frictionless joints. Frames cannot actually exists for there 
is always frictional resistance; however, many structures 
closely approach them, and whatever error there is may be 
considered somewhat as a factor of safety. 

A member of a structure which is in tension is called a tie; 
if under compression, a strut. 

If two or more members of a frame are connected at a 
joint, the stress in the members is considered as acting at the 
center of the joint; each member necessarily having a joint 
at each end to keep it in equilibrium, and obviously the stress 
in the member will be of the same amount throughout and 
will act along the axis, the effect on the joint at one end 
being in the opposite direction to the effect on the joint at 
the other end. 

If we consider a single joint of a structure, the resultant 
of the stresses in the members forming the joint must be 
equal in amount and opposite in direction to the load on that 
joint, for otherwise the joint could not be in equilibrium. 

All frames will be considered as loaded at the joints. In 
the case of continuous loads, such as the weight of a member, 
the equivalent parallel forces will be considered as acting at 
the joints, thus a uniformly loaded member would be con- 
sidered as having half the total load acting at each end. 

63. As a simple frame we will first consider a common tri- 
angular roof truss in which the weight of the roof itself forms 



Strength of Material. 



127 



the load, say iv lbs. per ft.-nin on each rafter. On the mem- 
ber AB there will be iva lbs. of which half will act at A and 
half at B. On the member BO there will be wl lbs., of which 
half will act at B and half at G; so that at B there will be 

a load of ^ (a + &) lbs. ISTow the supporting forces P and 

Q will have to sustain this whole load and their value is ob- 
tained just as we found it in the case of loaded beams; the 
moment arm of the load at B being the projection of the mem- 
ber AB on AC, or BC on AG; the lengths being found from 
our knowledge of distances, angles, etc. ^ow the lines of 



ifC***J 




Fig. 49. 



action of the load, ^-^ at A, and of the supporting force P 

are identical, but the directions of the forces are opposite, so 
we can use their difference for P without making any change 
in the stress of the members of the structure (this being the 
resultant of the two forces) and thus reducing the number of 
external forces to three. (It is convenient in calculating the 
supporting forces to ignore the load which acts on the lower 
end of the rafters.) We have now to find the stress in the 
members AB, BG, and AG, and we can do this (1) by cal- 
culating directly from our knowledge of the loads, angles and 
distances; (2) by means of the Eitter section, and (3) by a 
graphic method. Each of these methods will be explained in 



128 



Strength of Material. 



the three following articles, but each method must be known 
thoroughly, for with complicated frames it is necessary to 
employ more than one of the methods to get results. Before 
proceeding we will introduce a system of lettering which will 
be convenient, and which should be used in all cases. Ee- 
ferring to Fig. 50, the external loads at the ends of each 
member have been indicated by arrowheads, between each one 
of these external loads we place a capital letter, and in each 
space made by the members of the frame we place a small 




Fig. 50. 

letter To indicate any force we write or mention the letters 
on each side of it, for example, the left supporting force is 
EA the right one DE, the load at the peak BG, etc. The 
mUrnal forces will all have at least one small letter, though 
both may be small; thus the stress in the vertical member 
from the peak down is U, that in the member inclined to the 
left from the peak is Bl, the one inclined to the right Gc, 
etc With this method we letter the forces, which will be 
found most convenient when using the graphic method for 
finding their values. 



Strength of Material. 



129 



64. As an illustration, we will find the stresses in the mem- 
bers of a triangular roof truss of span 25 ft., rafters of 20 
and 15 ft. respectively, and which has a load of 100 lbs. per 
ft.-run on the rafters. The loads, additional dimensions and 
supporting forces as indicated in Fig. 51 are found by the 
methods of the preceding article, 

cos 6 = sin (^ 

sin (9 = cos (^ 



M 


= i 


« 


= f- 




1 




Pig. 51. 



For the equilibrium of the joint at the peak the sum of the 
vertical components of the stresses Aa and Ba must be equal 
to the load 1750, but must act in the opposite direction, 
therefore, 

Aa cos (f>-^Ba cos 6 = 1750, or, f Aa + ^Ba == 1750. (1) 

Considering the joint at the right support, the sum of the 
horizontal and of the vertical components of the forces acting 
there must be zero, or as before the vertical component of Ba 



130 Strength of Material. 

must equal Q, and the horizontal component must equal the 
stress Ca. Eesolving, horizontally, 

Ca = Ba cos <^, or, Ca = §Ba. (2) 

Vertically, 

Ba sin 4> — 1120, or, ^Ba = 1120. .' . Ba — 1400 lbs. 

Substituting in (1) 

iAa + ^. 1400 = 1750. .*. Aa = 1050 lbs. 

Substituting in (2) 

Ca = f . 1400 = 840. .-. (7fl^ = 840 lbs. 

^N'ow it is obvious that the stresses Aa and Ba will be com- 
pressive, and that Ca will be tensile, but in many cases it is 
not obvious, and with the graphic method will be shown a way 
of accurately distinguishing. Taking the joint at the peak, 
the external load acts down, so the resultant of the stresses 
Aa and Ba must act up; to do this they must push on the 
joint; a stress then which pushes on a joint is compressive 
and one which pulls is tensile. 

This method of finding stresses will hereafter be called the 
method by calculation, to distinguish it from Eitter^s method. 

65. E-itter's Method. — This may be called the method of 
sections, as it involves our fourth rule for the equilibrium of 
a structure. The moments may be taken about any axis and 
if possible we choose an axis about which the moments of all 
the unknown stresses cut by the section except one disappear. 
We will solve the problem of the preceding article by this 
method and to do so will first consider a section HH as shown 
in Fig. 52. If now we take moments about the joint at the 
peak, the moments of the stresses Aa and Ba will be zero and 
we will have to the left of the section only the external force 
P and the stress Ca to deal with; or to the right the external 
force Q and the stress Ca, there being no moment of the 



Strea^gth of Material. 



131 



1750-lb. load about the joint at the peak. Taking moments 
then the perpendicular distance from the peak to the stress 
Ca being 12 ft., and that to the line of action of the force P 
being 16 ft., we have 

630 X 16 = Ca X 12, or, Ca = 840, as before. 
Taking moments about the joint at the right support will 
eliminate the stress Ca, and still working with the forces to 
the left of the section we have 

630 X 25 = Aa X 15, or, Aa = 1050, as before. 




Care must be taken to use the perpendicular distance to the 
line of action of the forces. We have used 15 in this case 
because, it will be noticed, this frame is right-angled at 
the peak. 

To get the stress in Ba we must use another section, for it 
is clear that the only internal stresses involved are those of 
the members through which the section passes. We will use 
the section KK and take the moments about an axis through, 
let us say, the point L. (We may use any point.) Here 



132 Stejea^gth of Material. 

the perpendicular distance to the stress Ba is 7.2 ft., and to 
the supporting force § it is 9 ft. ; so 

1120 X 9 = 5a. X 7.2, or, Ba — 1400, as before. 
Instead of choosing L, for the axis of moments, we could just 
as well have taken the joint at the left supporting force from 
which the moments are 

1120 X 25 = 5a X 20, or, Ba = 1400. 

This method will be found convenient where the stress in 
particular members is desired, and is frequently necessary in 
the graphic method. 

66. The graphic method, invented by Clerk Maxwell, is 
based upon the principle of mechanics that a number of forces 
acting at a point are in equilibrium only when they can be 
represented in amount and direction by the consecutive sides 
of a closed polygon. In all frames the forces acting at the 
joints must be in equilibrium, and the line of action of the 
internal forces mu.st be in the direction of the axis of the 
member in which they are found. All the external forces 
taken in order around the frame will form a closed polygon, 
because by our first rule for the equilibrium of structures 
they must balance. Draw carefully a diagram of the frame 
and at each joint indicate by an arrowhead the load, if there 
be one; indicate also the supporting forces. This figure is 
called the frame diagram, and will be denoted by F. D. 

The diagram representing the polygons of forces acting at 
the joints is called the reciprocal diagram, and it will be de- 
noted by E. D. Having determined the am^ount and direc- 
tion of the external forces, plot them to a convenient scale, 
forming the external force polygon. In case the loads are all 
vertical, as in the example of Art: 63, the external force 
polygon is a vertical straight line, as in Fig. 53, the load 
1750 lbs. being represented by AB, the supporting force Q, 



Strength of Mateeial. 



133 



acting np, by BC, and P by CA, all of which forces have the 
same distinguishing letters in the E. D. as in the F. D. Now 
the stress in the left-hand rafter is in the direction of the 
rafter itself, so from A of the external force polygon draw a 
line parallel to this rafter to represent the line of action of 
the stress Aa, and from B draw a line parallel to the right- 
hand rafter to represent the line of action of the stress Ba. 
The intersection of these two lines will be the point " a " and 
if we connect G and a, the line will be found to be parallel 




Q = UZ.O 



Fig. 53. 




^:b 



to the tie rod of the frame, and Ca will represent the stress 
in the tie rod to the same scale as that used to plot the ex- 
ternal force polygon, just as A a and Ba will represent the 
stress in the rafters. This figure is called the Eeciprocal 
Diagram, and that the lines do represent the stresses may be 
proved, for the angles BAa and ABa are equal respectively 
to 4> and 9 of the F. D. by construction, and as AC is to scale 
equal to 630, Aa will equal 630 sec </> = 630 X | = 1050 ; 
Ba will equal 1120 sec ^ = 1120 X | = 1400; and Ca will 



134 Strength of Material. 

equal 630 tan </>, or 1120 tan 6, either of which will give 840. 
The results are then the same as those obtained by other 
methods. Of course, if the diagram is drawn accurately, the 
stresses may be measured off to scale. 

To find the kind of stress in any member, notice that the 
forces acting on the joint at the peak are the load AB, and 
the stresses Aa and Ba. If we pick these lines out on the 
E. D., they will form the polygon of forces (in this case a 
triangle) for the joint at the peak. Knowing the direction 
of one of these forces, we will indicate it by an arrowhead; 
for example, we know AB acts down, so we put an arrowhead 
pointing down on the line AB of the E. D. For the equi- 
librium of the joint at the peak the direction of the forces 
acting on it will be indicated if we suppose the arrowhead 
of AB to move in order around the sides of the polygon for 
this joint, starting in the direction in which it points. When 
on each side it will indicate the direction in which the stress 
for the corresponding member acts on the joint, and remem- 
bering that a push indicates compression and a pull tension, 
we can at once state what kind of stress exists in any member 
acting on the joint. The arrowheads are marked in the E. D. 
of Fig. 53 for the joint at the peak. They have been indi- 
cated for a single joint only, because if we consider any other 
joint involving the stress of any of the members acting on this 
one we would have two arrowheads on the same line pointing 
in opposite directions. This at first glance appears to be in- 
correct, but when we remember that the stress in any member 
acts in opposite directions on the joints at its two ends, and 
that we would now be working with a different joint, the 
apparent inaccuracy clears itself up; for example, if we con- 
sider the joint at the right-hand support, we will find the 
arrowhead for the stress Ba pointing toward B in the E. D., 
which we know to be correct, as the right-hand rafter is in 
compression and therefore pushes on this joint. 



Strength of Material. 135 

Examples : 

1. The slope of the rafters of a simple triangular roof- 
truss is 30°. What is the stress in each member when loaded 
with 250 lbs. at the peak? 

Ans. Eafters, 250 lbs.; tie rod, 216.5 lbs. 

2. A beam 15 ft. long is trussed with steel tension rods 
and a strut at the middle forming a simple triangular truss 
2 ft. deep. What is the stress on each member when loaded 
with 2 tons at the middle ? 

Ans. Strut, 2 tons; tension rods, 3.88 tons; thrust on the 
beam, 3.75 tons. 

3. A small brow, 6 ft. broad and 20-ft. span, carries a load 
of 100 lbs. per sq. ft. of platform. It is supported by two 
simple triangular trusses 3 ft. deep. Find the stress in each 
member. 

xAqs. Strut, 3000 lbs. ; tie rods, 5220 ; thrust on beams, 
5000 lbs. 

4. The rafters of a simple triangular roof -truss slope 30° 
and 45°; span 10 ft., the rafters are 2 J ft. apart, and the 
roof weighs 20 lbs. per sq. ft. Find the stress on each 
member. 

Ans. Stress on tie rod, 198 lbs. 

5. A small brow, 4 ft. broad and 20-ft. span, carries a load 
of 100 lbs. per sq. ft. of platform. A load of 1 ton passes 
over the brow; what is the stress in the members when this 
load is in the middle of the brow ? 

Ans. Compression of strut, 3120 lbs. 

6. The tie rod of a simple triangular roof -truss is 25 ft. 
long and inclined at 30° to the horizontal. The supports are 
at its ends and the rafters from its ends are 20 and 15 ft. 
long and are loaded with 30 lbs. per ft.-run. What is the 
stress in each member? 

Ans. Stress in rafters, 148 -j- lbs. and 512 -f lbs. ; tie 
rod, 174 -f lbs. 



136 



Strength of Material. 



CHAPTEE XIII. 

Framed Structures — Continued. 

67. A roof truss with a vertical member from peak to tie, 
has flooring laid on the horizontal ties, in addition to the load 
on the rafters, so that when divided at the joints the load 
will be as shown in Fig. 54. 

4 




Fig. 54. 



Proceeding in the usual wa}^ we get the E. D. as shown, 
and from it 

Aa^ 10 tons C. (compression). 
Bh — 8V2 tons C. 
ha = 3-J tons T. (tension) . 
Da=:Oh = S tons T. . 

To get the stress Aa, for example, by the method of sections : 
taking a section HH, and moments about the joint at the 



Strej^gth of Material. 



137 



riglit-hand support, iising forces to the left of the section, 
we get, X being equal to 16.8 ft., 

P X 28 = Aa X ^, or, Aa = 10 tons, as before. 

The method by calculation is seldom used as compared to the 
other methods it is complicated. 

68. A "King-Post Truss,'^ slope of rafters 45°, and having 
struts to the middle points of the rafters as shown in Fig. 55, 
has a load of 8 tons uniformly distributed on each rafter. 




7?.-27. 



Fig. 55. 



Find the stress in the members. The loads are as shown in 
the F. D. The E. D. gives us : 

Aa = Dd = 6V2 tons C. 

Bb = Cc = 4V2^tons C. 

ah = cd = 2V2'tons C. 

Ea — Ed — Q tons T. 

5c = 4 tons T. 
To get the stress Ed, for example, by method of sections, 
take section HH, moments about joint at peak, calling span S. 

Q X^ = Edxi- .'. Ed = 6 tons, as above. 



138 



Strength of Material. 



69. In the preceding example^ suppose the roof had also to 
sustain a horizontal wind pressure of 6 tons, nniformly dis- 
tributed on the right-hand rafter. Putting in all the loads 
they would be as shown in Fig. 56, and the walls would have 




Fig. 56. 



to sustain in addition to the vertical load a lateral pressure 
of 6 tons. The resultant of the loads is not now vertical, 
therefore, our supporting forces will be inclined. Obviously, 
the effect is greatest on the left-hand support, so the inclina- 



Strength of Material. 139 

tion of P will not be the same as that of Q. Calling the 
angles that P and Q make with the vertical, <^ and 6 re- 
spectively, and the span 8, and taking moments about the 
joint at the left support (Fig. 56, a), 

55cos^ = |x4+|x4 + |^X4 + 3>S'-|x3-|xi.5. 

.'. gcos^ = 6.5, 
and, b}^ the same method, 

P coscf) ^ 9.5, 

which shows the sum of the vertical components of the sup- 
porting forces to be equal to the vertical loads as it should 
be, also 

P sin <^ -|- ^ sin (9 = 6 tons, 

or the horizontal components equal the horizontal loads. Pro- 
ceed now with the external force pol3^gon to and including 
the force HI. The next two forces are the supporting forces, 
we know the external forces give us a closed polygon, there- 
fore, the resultant of the two supporting forces is the dotted 
line I A. We know the sum of the horizontal components of 
the two supporting forces is 6 tons, so we will draw a line 
parallel to the direction of AB, BC, etc., at a distance from it 
equal to 6 tons as per our scale. We also know the vertical 
component of the force P is 9.5 tons, so laying off from A, 
toward D the distance 9.5 per scale, if through this point we 
draw a horizontal line, the point J must be somewhere on it, 
and also somewhere on the first line. Therefore, J is located 
at their intersection. Connecting J and I we find the line is 
vertical and therefore that the supporting force at Q is ver- 
tical and equal to 6.5 tons. Connecting J and A we have 
the supporting force P both in magnitude and direction, and 
find it equal to 11.236 tons and at 32° 16' 32" with the 
vertical. 



140 Strength of Material. 

Of course practically the rigiit-hand support would sustain 
some of the horizontal load, due to friction and the way the 
roof is secured to the walls, but if the left-hand wall is built 
strong enough to take all the horizontal effect our structure 
will be safe. As we can always resolve any oblique force hori- 
zontally and vertically, we may assume that the supporting 
wall, on the side from which the horizontal component comes, 
has a roller on its top so that the su.pporting force on that 
side will have to be vertical. 

Proceeding now, we complete the F. D. by putting in the 
supporting forces, then finish the E. D. as shown in the figure 
and find the stresses to be as follows : 

Ba — 7.5 V 2 tons C. Ec = 4V2 tons C. 

Ch — 5.5 V^ tons C. he = 5.5 tons T. 

Gd = 4.5 V2 tons C. cd — 3.5 V2 tons C. 

la = 2 V2 tons C. Jd — ?, tons T. 
Ja = 1.5 tons T. 

To get the stress cd by the method of sections, we must know 
the stress Jd, then take a section KK and use the forces to 
the right of the section taking moments about the joint at the 
peak. For Jd, using the section HE, 

JdX-^ =6.5 X-f--^ X^- — 1.5 X-|^, 

or, 

Jd = 3 tons . T. ; 
now, using section KK (a section through Ec, cd and dJ), 

+ 2 X 4 + 4 X 4 + ^ >< T 
from which cd — 3.5 V2'. C, as before. 



Strength of Material. 



141 



70. An inverted '' Queen-Truss " is loaded, as sliown in the 
F. D. of Fig. 57, with unequal loads over the two struts. 
With this loading the E. D. is as shown. The stress in the 
diagonal is tension. If the loads had been reversed, i. e., 2W 
for BC, and W for AB, the stress in the diagonal would have 
been compressive. If the diagonal member had been omitted, 
the E. D. would not have been a closed figure (except in the 
case of equal loads over the struts), showing that the frame 



2W 




1 


I 


^^ 




c 





ni). 



B 



Fig. 57. 



would not then be in equilibrium. The diagonal member is 
then necessary in bridges of this kind, because of the travel- 
ling loads they must carry. 

71. A Bollman truss is 24 ft. long, 3 ft. deep and carries 
a uniform load of 3 tons per ft.-run. 

The F. D. is as shown in Fig. 58. Proceeding with the 
E. D. we draw the external force polygon, the direction of the 
stresses Aa, Bd, Cf,.Dg, De, Be, and I)}). Here we have to 
stop for we have none of the points represented by the small 
letters. The simplest way out of our difficulty will be to find 
9 



142 



Strength of Material. 



the amount of stress in one of the members by the method of 
sections. Taking the section EH, moments about the point 
where the two long tension members cross (marked with 
circle) and with the forces to the left of the section, 

5^ X f= 24 X 12 — 24 X 4, or, Bd = 85* tons. 

Laying this value off to scale on the line Bd gives us the 
point d, after which we have no difficulty in obtaining the 
E. D. shown. 




P.^y 



G^-^¥ 



r.D. 




n.u. 



Fig. 58. 



72. The force diagram of Fig. 59 shows an i\^ girder of 
four panels with a uniform load over the lower platform. 
Having drawn the external force diagram the stress l<l(h being 
horizontal, the point a must lie somewhere on a horizontal 
line through B ; the stress Aa being vertical the point w must 
lie somewhere on a vertical line through A. Therefore the 
point a coincides with the point B of the E. D., or the stress 
Ba is equal to zero with this loading. The member is neces- 
sary to the bridge, however, for even if it did not support part 
of the platform it would be required- to keep the frame from 
turning about the upper joint over the left support. The 
stress Bli also proves to be equal to zero in the same way. 
Proceeding now we get the E. D. as shown and finding the 
stress ed is also zero. In this case, too, the member is neces- 



Stren^gth of Material. 



143 



sary, for with a joint at the middle of the top boom the 
stresses Ad and Ae being compressive would cause it to turn, 
and without the joint the necessary length of the member 
would cause it to buckle. Having now the R. D. the stresses 
in the members are easily obtained. In designing N girders 
we must consider the effect of travelling loads, for the bridge 
in addition to sustaining its own weight must be strong 
enough to sustain the stresses caused by the loads which cross 
it. The method of constructing bridge platforms (the planks 



T.^. 




Fig. 59. 



or railroad ties being placed across stringers between joints) 
puts practically all the bending stresses on the horizontal 
parts and all the shearing stresses are sustained by the diag- 
onals (the shearing stress being equal to the vertical com- 
ponent of the stress in the diagonal). Therefore, consider- 
ing only the weight of the structure, the diagonals at the left 
of the middle are inclined to the left and those to the right 
of the middle to the right, because those directions put them 
in tension. (For the same load a rod in tension requires less 
cross-sectional area than one in compression, hence less ma- 
terial, less weight and less cost.) If we consider a bridge of 
this kind with a concentrated load moving across it, referring 
to Fig. 41, Art. 50, the load entering the bridge at the left 



144 



Strength of Material. 



end^ we see that positive shearing force predominates at the 
left end and negative at the right, the curve in this figure 
being for the travelling load only. Therefore the total shear- 
ing force due to loth the weight of the strnctnre and the 
travelling load, will change sign from plus to minns at some 
position of the travelling load other than the middle of the 
bridge. Obviously, this will change the hind of stress in the 
diagonal at this point and the dimensions of the diagonal 
here must be made such as to sustain the new stress, or we 
must put in a second diagonal inclined in the opposite direc- 
tion. The latter method is used because of the increase of 
weight necessary with a diagonal under compression, and also 
because a tension rod is less expensive and quite as efficient. 
These extra diagonals are called counter graces, and the 
above is the reason for counter bracing some of the middle 
panels of iV girders. 

73. A Warren girder, angle between members 60°, carries 
a uniform load on the lower platform as shown in Fig. 60. 




The R. D. is as shown, and the stresses in members are easily 
obtained from it. The remarks of the preceding article on 
counter bracing apply equally to this girder. If we draw 



Strength of Material. 145 

vertical lines through each joint and call the part between 
two vertical lines a panel^ the calculation will be Just the same 
as for the N girder. 

Examples: 

1. A king-post truss^ slope of rafters 45°^ has struts to 
the middle point of the rafters as in Fig. 55. AB ^ 2w; 
BC = 3w; and CD = 4w. Find the stress in each member 
by the method of sections and also from the E. D., stating 
the kind. 

2. A bridge is constructed of a pair of Warren girders with 
the platform on the lower boom^ which is of four divisions. 
40 tons is uniformly distributed over the left half of the 
bridge. Find the stress in each member, stating the kind. 

3. A bridge, 60-ft. span, is supported by a pair of N girders 
of 6 panels, height 8 ft. The platform is on the upper boom 
and carries a uniformly distributed load of 12 tons on the left 
half of the bridge. Find stress in each member, stating the 
kind. 

4. A roof is constructed of two rafters at right angles, a 
horizontal member connects the m'i^dle points of the two 
rafters, the ends of this member are connected with lower ends 
of the opposite rafter. The joints of the peak and the middle 
of each rafter carry a load of 1 ton. Find the stress in each 
member and the kind, if the span is 20 V 2 ft. 

5. A bridge of 60-ft. span is constructed of a pair of War- 
ren girders, the upper boom, which supports the platform, is 
of 6 divisions, the lower boom of 5. The platform is also 
supported by struts from the joints of the lower boom, and 
carries a load of f ton per ft.-run. The supporting forces 
are at the ends of the upper boom. Find the amount and 
kind of stress in each member. 



146 Strength of Material. 

6. A Bollman truss of 48-ft. span and 12 ft. deep carries 
a uniform load of J ton per ft.-nin over the left two-thirds 
of the horizontal boom. Find the amount and kind of stress 
in each member by the method of sections and also from 
the E. D. 

7. A bridge ;, 96-ft. span, is supported by a pair of N girders 
with 8 panels 9 ft. deep. The platform rests on the upper 
boom and is loaded with 96 tons uniformly distributed. Find 
the amount and kind of stress in each member. 

8. A concentrated load of 30 tons is to pass over the bridge 
of example 7. What panels should be counter braced? 

9. A Warren girder, 9 ft. long, projects from a wall. The 
top boom is of three divisions, the lower of two, but has a 
horizontal strut from the inner joint to the wall. With a 
load of 2 tons at the outer end of the upper boom find the 
stress in all the members, stating the kind. 

10. Suppose the load of example 9 at each of the other two 
joints of the upper boom, and thence deduce the results for 
a distributed load of f tons per ft.-run. 



Strength of Material, 



147 



CHAPTEE XIV. 

Framed Structures — Continued. 

74. An example of a scissors-beam truss is shown in Fig. 
61. With vertical loads the E. D. shown is readily obtained. 




T>»%w 



Fig. 61. 



With all roof-trusses the loads due to wind pressures must be 
taken into account. We have seen that this may be done by 
resolving the wind loads into their horizontal and vertical 
components (Art. 69) and considering the total effect as sus- 
tained by one of the walls. It may also be done by combining 
the loads on each joint and using the resultant load on each 
joint in drawing the external force diagram, or by finding the 
E. D. for the vertical and inclined loads separately and alge- 
braically, adding the stress due to each for the total stress in 
the members. 

75. When the external loads and the members of a struc- 
ture are all in the same plane we can always find the stress in 



148 



STREj;rGTH OF Mateeial. 



the members by means of the reciprocal diagram, for example, 
the F. D. of Fig. 62, shows a common crane carrying a weight 
W. The external forces are as shown, the forces BC and DA 
being obviously necessary to prevent overturning, their value 
being readily found by taking moments. The force CD is 




Fig. 62. 



equal to the weight. The external force diagram is a rect- 
angle. The forces CD and DA can be combined if desired, 
their resultant being equal to the dotted line CA. The R. D. 
is readily completed as shown, and from it the stresses are 
easily found. Cranes are usually arranged so that the load is 
lifted by means of a tackle, the hauling part of which leads 
down to a drum secured somewhere along the crane post. 



Strength of Material. 



149 



W 



The stress in this rope (which is equal to — , where n is the 

number of parts of rope between the two blocks of the tackle) 
augments the compression in the jib and reduces or increases 
the tension of the jib stay, depending on whether the drum is 
above or below the point where the jib joins the crane post. 




These additional loads on the members must be taken into 
account. This can be done by introducing an additional 
external load at the end of the jib, acting in the direction of 

W 



the hauling part of the tackle and equal to 



This addi- 



tional load will cause the supporting force DA of Fig. 63 to 
be somewhat inclined. 

To draw any external force diagram we must know all but 



150 Strength of Material. 

one of the external forces. In this case we know only the 
weight lifted and the tension of the rope. We can, however, 
find the force CD by taking moments about the lower end of 
the crane post, as 

W 

whence 

ny 
The force DA is the resultant of a horizontal force similar 
to DA of Fig. 62 and the slightly inclined supporting force. 
Except the resultant now we know all the external forces, so 
our external force polygon is ABGD, the line DA representing 
the amount and direction of the supporting force. The E. D. 
is now readily found to be as shown, and the stresses are 
easily obtained. 

76. Incomplete Frames.— A frame having just enough 
members to enable it to retain its shape under any kind of 
loading is complete. Frames may have more members than 
necessary, under which circumstances some of the members 
are redundant, but frames without a sufficient number of 
members to enable them to retain their shape under all cir- 
cumstances, with any kind of loading, are said to be incom- 
plete. When frames are incomplete we will find the recipro- 
cal diagram for different loadings giving intersections for the 
same point in two or more places; as, for example, we know 
the mansard roof-truss^ shown in the F. D. of Fig. 64, is 
incomplete because there is nothing to prevent the three upper 
joints from turning. Suppose TT^ = W^ and drawing the 
E. D., we find the point " a^^ falls in two different places on 
the horizontal line from E ; this is clearly impossible, for the 
stress Ea cannot have two values. The frame then is not in 
equilibrium. If W^ had not been equal to W^ the two a's 



iSteength of Material. 



151 



would not have been on the same horizontal line. We could 
arrange the loads so that this frame would appear from the 
E. D. to be complete ; as, assume either of the points " a " to 
be correct, say the right one, and from it draw lines (dotted 
in figure) parallel to the stresses Ba and Ca. These lines will 
intersect the external force diagram at points indicating loads 
which would put the frame in equilibrium, but the first puff 




Fig. 64. 



of wind would change the loading and cause the frame to 
collapse. 

To make this frame complete, at least two additional mem- 
bers are necessary, for there are three joints which are likely 
to turn. 

The two additional members, shown in the F. D. of Fig. 65, 
will complete the frame, as is shown by the E. D. being a 
closed figure and possessing no duplicate points, no matter 
what loads are applied. 

Neither of these additional members alone will do this, as 



152 



Strength of Material. 



will be made clear if we discard one of tliem, vary the loads 
and draw the E. D/s. 

A frame, then, to be complete, must remain in equilibrium 
under any loads which do not stress its members beyond the 
elastic limit. 




Fig. 65. 

77. We have been considering in all these structures the 
stress which acts along the axis of the member. When mem- 
bers are not straight, or when they carry loads at points other 
than the joints, the stress on them is not simply a thrust or 
pull in the direction of the axis, but includes a bending and 
a shearing action. Fig. && shows a single member of a struc- 
ture with two intermediate vertical loads. If we resolve all 
these forces along and perpendicular to the member, the com- 
ponents perpendicular to the member will cause bending and 
shearing stresses in it, and using these components, W cos 6 



Strength of Material. 153 

and Wi cos as the loads^ and P cos ^ and Q cos ^ as the sup- 
porting forces^, we find the shearing and bending action just 
as we did in Chapters VI and YII. 

These intermediate loads cause also a stress along the axis 
of the member and this stress (the thrust T shown) between 
W and W^ will be equal to Wj^ sin 6 and between P and W it 
will be the sum of T^^ sin and W sin ^, or ( W + ^i) sin 0, 
which value is the reaction of this member on the one next 
below it. 



Fig. 66. 

Obviously {P -{- Q) sin 6 causes an equal reaction on the 
member next above it. 

Just here we can show that our assumption that the loads 
act on the joints is correct, for the action on these joints of 
the structure, considered as a whole, caused by this member is 
equal to a force P acting dowmvard at the left end, and a 
force Q acting downward at the right end; and obviously 
without error we can use these forces P and Q (reversed) in- 
stead of considering the separate loads. When a member 
carries a uniform load, as is usually the case, the values of P 
and Q are each half the load and the thrust T is the value of 
the axial stress at the middle of the member. We have not 



154 Strength of Material. 

considered the bending and shearing stresses in members of a 
structure until now, because they are comparative!}^ small, a 
compression member being necessarily large to sustain the 
thrust due to all the other members which affect it, will be 
little inconvenienced by the small stress due to the bending 
and shearing caused by its own load, and the tension mem- 
bers, in properly built structures, will have only their own 
weight to cause bending and shearing. 

Examples: 

1. A distributed load of 4 tons is carried by a scissors 
truss whose rafters, 12 ft. long, are at an angle of 90°. Two 
of the three other members join the lower end of our rafter 
with the middle of the other. The third is horizontal and 
joins the middle points of the rafters. Find the ptress in all 
the members. 

2. In example 4, Chapter XII, find the shearing force and 
bending moment and thrust for each point of each rafter and 
draw curves showing results. 

3. A simple triangular frame, sides 3, 4 and 5 ft. long, has 
the 5-ft. side horizontal. Its numbers weigh 10 lbs. per 
ft.-run, and the inclined sides each have 50 lbs. at the middle. 
Draw curves of thrust, shearing force and bending moment 
for each member. 

4. A suspension bridge carries a platform 8 ft. wide, span 
63 ft. suspended by 6 equidistant tension rods. The lowest 
joint is 7 ft. below the highest, and the cable is formed of 6 
straight members. It carries a load of 1 cwt. per sq. ft. of 
platform. Find the sectional areas of the cable members, 
allowing a stress of 4 tons per sq. in. for the material. Is 
this frame complete ? 

Ans. From left end, 3.72, 3.6, 3.5 sq. in.; middle, 3.47 
sq. in. 



Strejstgth of Material. 155 

5. A mansard roof -truss, as in Fig. 65, carries 1, 2 and 
3 tons on the left, top and right joints. The rafters make 
angles of 45° and 30° with the horizontal. Find the stress 
in all the members. 

6. A crane, as in Fig. 62, has the tension members Ac 
horizontal and 6 ft. long. The members he, ha and aC are 
respectively 8, 6 and 4.5 ft. long. Find the stress in each 
member when supporting a weight of 12 tons. 



156 



Streis^gth of Material. 



MISCELLANEOUS PEOBLEMS. 

78. Reinforced Concrete Beams. — Concrete is much used in 
building at present and though its tensile strength is ver}^ 
low compared to other materials, the ease with which it is 
handled, transported and shaped, together with the fact that 
when reinforced with steel its tensile strength is satisfactor}^ 
and the structure is non-combustible has made it most accept- 
able to builders. The determination of the constants for 
concrete is rather difficult considering the changes they suffer 




Fig. 67. 



due to the different kinds of cement used and the different 
methods of mixing. Strength tests for the cement itself 
vary from 40 to 1000 lbs. per sq. in. for tension and from 
1100 to 12,000 lbs. per sq. in for compression. 

In all building the endeavor has been to subject the con- 
crete to little or no tensile stress, but to have all that stress 
sustained by the steel reinforcement, allowing the concrete 
to sustain as much as possible of the compressive stress. 
Tests have shown the compressive strength of concrete to 



Strength of Material. 157 

vary from 750 to 5360 lbs. per sq. in. and the modulus of 
elasticity to range from 500,000 to 4,167,000 in in.-lb. units. 
A method of finding the moment of inertia of the section 
of a reinforced concrete beam is as follows : Assuming that 
the elongation of the steel reinforcement is the same as for 
the concrete at equal distances from the neutral plane which 
will be true if the concrete adheres closely to the metal rein- 
forcement, we have by Hook's law, 

Psteei — Esteei X extension 
and 

P concrete — — J^ concrete /\ CXtenSlOn, 

or, 



1^ -1^ 
E.- E^ 



from which 



P. = Jp. 



Fig. 67 represents the section of a concrete beam rein- 
forced by three steel rods on the tension side of the neutral 
plane, N". P. If now^ we suppose the beam to be entirely of 
concrete and to retain the same strength and the same depth 
we will have to broaden the section in wake of the steel rods 
to conform with the fiber stress which that part of the section 
can now withstand. If y is the distance from the neutral 
axis to an elementary area, dA, of the steel, the moment of 
the stress about the neutral axis will be 

yp.dA = ypc -^ dA. 

So, if the section is to be considered homogeneous (all con- 
crete), w^e must have the areas vary as the modulii of elas- 
ticity or the area of concrete which is equivalent to the present 
area of steel will be as Es is to Ec • The equivalent concrete 
10 



158 



Steexgth of Material. 



section will therefore be something like that shown in Fig. 68. 
The general equation of bending gives for the fiber stress 

V J ^ 

I being the moment of inertia of this equivalent section. For 
example, suppose the beam of Fig. 67 to be 6 in. wide, 10 in. 
deep and the reinforcing rods to be ^ in. square with their 
lower faces 1 in. from the surface of the concrete. Let Es be 
to Ec as 15 to 1. Here the area of the steel section is f sq. 
in., hence the equivalent area of concrete is 

15X5 =¥sq. in., 



Fig. 68. 



and as the rods are half an in. deep the additional breadth at 
this part of the section due to the steel will be ^ ins. But 
there is at this part of the section beside the steel f ins. of 
concrete, so the ti;]ioIe width of the section at the point will be 

45 , 9 54 . 

^+¥ = "2 ^"^-^ 

and the depth of the broadened part will be the same as the 
steel, or -J in. We have now all the dimensions of the equiva- 
lent concrete section from which we find that the neutral axis 
is close to 4.4 ins. from the bottom of the beam, and the 
moment of inertia of the section about this axis is about 626 
in in. -units. 



Strength of Material. 



159 



It has been shown by experiment that small cracks appear 
on the tension side of a reinforced concrete beam as soon as 
the fiber stress reaches the limiting tensile strength of the 
concrete. From this fact it is clear that practically all of 
the tensile stress is supported by the steel reinforcement. 
With this assumption then, referring to Fig. 69, the sum of 
all the tensile stresses on one side of a section must equal the 
sum of all the compressive stresses on the same side of the 
section. If then A is the known area of the steel section 




Fig. 69. 



and Ps the stress in it, and hx the area of the concrete sec- 
tion on the compressive side of the neutral plane and pc the 
maximum compressive stress in it, we have, assuming the 
compressive stress to vary directly as the distance from the 
neutral plane 

Aps = ipcl)x. (1) 

We know the position of the steel reinforcement, and calling 
its distance from the top of the beam h, we have the sum of 
the moments of the stresses about the neutral axis equal to 
the bending moment, or 

Aps{h — x)-{- ipctx . fa; = i¥. (2) 



160 Strength of Material. 

'Now as the bending is considered uniform the extension of 
the steel on the tension side must equal the contraction of 
that part of the concrete on the compression side which is at 
the same distance from the neutral axis as the steel, therefore, 
by Hook's law (strains vary as distance from neutral plane) 

We now have three equations from which we can find the three 
unknown quantities i[)s , pc and x; ps and pc being the total 
stresses. If we have a reinforced concrete column under a 
load, the parts of the load borne by the concrete and steel can 
be found because the change of length of the two materials 
must be the same. We have, therefore, the equations 

A,E,-A^E,^ ^^^ 

and the load 

L = ps + pc, (2) 

where I is the length of the column. As and Ac the sectional 
areas of steel and concrete respectively. Concrete columns 
are usually made so short that bending does not enter into 
the calculations, which is the case when the column is not 
longer than twelve times its least diameter. A difficulty in 
investigating the bending of reinforced concrete beams lies 
in the fact that the modulus of elasticity of concrete under 
compression decreases somewhat as the loads increase. This 
will evidently cause an upward movement of the neutral 
plane. Several theories with different assumptions have been 
advanced but the investigation of these beams may be con- 
sidered to be still in an experimental stage. The use of the 
equivalent section gives fair results. Another theory assumes 
the stress in the concrete to vary as the ordinates of a para- 
bola and that the modulii of elasticity for the different 
stresses are those found by tests for varying pressures. 



Strength of Mateeial. 161 

79. Poisson's Ratio. — Experiment shows us that when a 
piece of material is stretched the area of a transverse section 
is reduced somewhat in addition to the change made in the 
length of the piece. This change of cross-sectional area is 
quite marked if we stretch a piece of rubber because the elon- 
gation is great, but with materials such as are used in build- 
ing the change of cross-section is scarcely perceptible for the 
elongation is very slight. Tests show that the lateral expan- 
sion or contraction due to compression or tension bears a 
constant ratio to the change of length. This is known as 
Poisson^s ratio. If then the extension per unit length of a 
round rod under tension is e, the length of its diameter d 
will be reduced an amount equal to hed where h is the value 
of Poisson's ratio. Poisson's ratio is for 

Steel 297 

Iron 277 

Copper 340 

Brass 357 

If a rod of length I, and of unit square section is under a 
tensile stress F along its axis and a compressive stress R, per- 
pendicular to the axis on two opposite faces, the total exten 
sion in the direction of the axis is that due to the tensile 
stress plus that due to the compression stress, or, 

, , T , • Fl kRl 

total extension ^ ^r 4- — cr— . 

80. Stress in Guns. — We found in Art. 10 that the hoop 
stress of a thin cylinder under the internal pressure of s lbs. 

per unit area was p ^ -r- where t was the thickness of the 
z 

metal and r the radius of the cylinder. In this case the 
metal was considered so thin that the stress might be taken 
as uniform throughout the section. If, however^ the metal 
is tliiclz we cannot use this formula for the stress will not be 



162 



Strength of Material. 



nniform throughout the section but will vary in some way 
with the distance from the axis of the cylinder. 

Let us consider a closed cylinder of internal radius r^ and 
external radius r^ under an internal pressure ;S' and an ex- 
ternal pressure 8^ per unit area. The cylinder having closed 
ends there will obviously be a longitudinal stress which is 
readily seen to be the total pressure on an end divided by 
the sectional area of the cylinder. We will call this longi- 




FiG. 70. 



tudinal stress T and will assume it is uniform throughout 
the section. If . now we consider the forces acting on any 
element of volume within the material of the cylinder the 
element will be in equilibrium under the action of the longi- 
tudinal stress T, the hoop stress H and the radial stress E. 
From the preceding article we have for a value of the total 
extension at the element in the direction of the axis of the 
cylinder 



Strength of Material. 163 

Tc being the value of Poisson^s ratio for the material. ISTow, 
if sections remain plane while stressed, we may assume the 
total extension due to these three forces constant, and T being 
constant, we must have H plus R equal to a constant, 

E + R = C. (1) 

Let Fig. 70 represent a right section of the cylinder of unit 
thickness (perpendicular to the paper) and let the circular 
element shown be of width dr. If we call the radial stress 
at the inner surface of the element R, the hoop stress at the 
inner surface of the element will be (Art. 10) equal to Rr. 
The radial stress at the outer surface of the element will be 
R -\- clR and as the radius here is r -{- dr the hoop stress at 
the outer surface of the element will be (R -\- dR) {r -\- dr). 
The hoop stress over the sectional area of the element will be 
{R -{- dR) {r -]- dr) — Rr. But we have called the hoop 
stress at any point H, therefore, the hoop stress on this ele- 
ment will be H multiplied by the sectional area of the ele- 
ment, hence we will have, the element being of unit thickness, 

{R + dR) (r + dr) —Rr = Hdr; 
or, neglecting the higher order of infinitesimals 

rdR-^Rdr = Hdr. (2) 

From (1) we have H^C — R and, substituting, 

rdR + Rdr — Cdr — Rdr, 
multiplying both sides of the equation by r, we have 

r^'dR -f 2rRdr = Crdr, 
or, 

d{r^R)^Crdr, 
and, integrating, 

where C^ is the constant of integration. 



164 Strength of Material. 



From the last equation we get 



and, substituting this value of E in ( 1 ) , Ave get 

'Now the cylinder has an internal pressure of S lbs. per unit 
area, and an external pressure of 8^^ lbs. per unit area, so if 
r = r^ we have R at the inside surface equal to S and if 
r = 7\ we have R at the outside surface equal to ^S^^ and, sub- 
stituting, we get 

and 

5.= f+|f; (6) 

solving (5) and (6) we have 

^1— ^2 y.2 > {,() 

and 

C- ^TZZT^' (8) 

substituting the values from (7) and (8) in (3) and (4) we 

and 

If there is no external pressure as in the cases of cast guns 
and hydraulic pipes, 8^^ will be equal to zero and our formulas 
become 

and 

ff=-J^'(^ + A- (12) 



Strength of Material. 165 

From equations (11) and (12) we see that the greatest stress 
is at the inner surface of the C3'lincler and here the total ex- 
tension found by the method of Art. 79 must not exceed the 
elastic limit of the material. It will be noticed that the 
values of H and R found above do not take into considera- 
tion the lateral contraction of the material mentioned in the 
preceding article. If we consider this lateral contraction and 
call H^ the stress which would produce an extension equal to 
the total elongation found by the methods of Art. 79 we 
would have 

H^ = H — hR — IT. 

For the material used in the construction of the modern 
naval gun the value of Poisson's ratio is ^. Putting in this 
value for h and the values of H and E from equations (9) 
and (10), the value of T being found directly from the ex- 
ternal and internal end pressures and the sectional area of 
the cylinder to be 

we have, after reducing, for the true hoop stress 

^1- r,' — r,' ^ 'Sr'{r,' — r.^) ' ^^'^^ 

which is the formula used for the hoop stress for naval guns. 

81. Built-up Guns. — The guns at present used in the navy 
are known as " built-up '' guns ; that is, they are built of 
several pieces which are made separately. The inner piece 
is called the tube, outside of it is the jachet, and outside of 
the jacket are the hoops. The exterior diameter of the tube 
is made a little greater than the interior diameter of the 
jacket, the exterior diameter of the jacket a little greater 
than the interior diameter of the first hoop, etc. The process 
known as '' assemhling " consists in heating the jacket until 



1*56 Strength op Material. 

it expands sufficiently to just slip over the tube, after which 
It IS a lowed to cool. In cooling the jacket contracts and 
grips the tube firmly, putting considerable external pressure 
on It. The hoops are then put over the jacket in the same 
way. This way of securing the several parts together is 
called the method of hoop shrinkage, and obviously if two 
parts are assembled by this method the inner one will be 
under hoop compression and the outer one under hoop tension 
Considering a cylinder formed of two parts, let the radii 
after assembling be r, , r, and r, , the least radius being r 
Let i?3 , E^ and R, be the radial stresses at the inner, inter- 
mediate and outer surfaces. We will call the difference be- 
tween the inside radius of the jacket and the outside radius 
of the tube Ufore assembling e, then if after assemblino- e 
IS the decrease in the length of the radius of the tube and / 
the increase of the length of radius of the jacket we have 
« = «2 + ei. (1) 

If H^ is the hoop compression on the outside surface of the 
tube and H^ the hoop tension on the inside surface of the 
jacket, e, the change of length of the radius of the tube due 
to the stress H, will be J^ (Chapter I) and e, the change 
of length of the radius of the jacket due to the stress H^ will 
be — jr^, hence 

" - E ^^T"' 
or, 

^. + ^.= f. (2) 

There being no internal pressure in this cylinder, equation 
(13) of the preceding article will give us a formula for the 
hoop compression at the outer surface of the tube if we let 



Strength of Material. 167 

S ^ 0, /Si ^ i^2 ^^^ ^' = "^'2 ^^ ^l^^e same time changing r^ of 
equation (13) to 7\ and r^ to r, . This gives us the equation 

^' - r/ — n^ ^ 3r/ (r/ — r3^) ' ^ "^^ 

There being no external pressure on the Jacket equation (13) 
of the preceding article will also give us a formula for the 
hoop tension at the inner surface of the jacket if we put 
S-^=i 0, S ^ R2 and r =z r^ . This gives us the equation 

TT ' 2 -'^^2 ^' A ' 2 -'■^2 ^A\ 

^1- r,' — r,' dr,'{r,' — r,')' W 

Taking the factor R2 from the right side of equations (3) 
and (4) all the quantities that remain are known (they being 
the different radii). Calling the value of these remaining 
quantities in equation (3) C^ and those for equation (4) Co 

''^'^^^' H^^E.C,, (5) 

^"""^ H^ = — R,C,, (6) 



dividing (5) by (6) to eliminate R2 we have 
H, — C, • 



(7) 



Solving the simultaneous equations (2) and (7) for H-^ and 
Ho gives 

_ C,Fe 
^^-(67,-6'Jr/ ^^^ 

and 

Equations (8) and (9) give the hoop stresses at the joint 
and having these we can from either of the equations (3) or 
(4) find the value of R^ . Having R2 equation (13) of the 
preceding article will give us the value of the hoop com- 
pression at the inner surface of the tube at which place the 



168 Streis^gth of Material. 

lioop compressive stress is obviously greatest when the gun is 
at rest. At the instant of firing a gun the internal pressure 
becomes very great and both the tube and jacket will then be 
under hoop tensile stress^ but obviously the first effect of the 
internal pressure must be to overcome the initial compressive 
stress in the tube, though it increases the tensile stress in the 
outer hoops where the effect of the internal pressure is least; 
thus the built-up gun can sustain greater internal pressures 
than the solid gun. 

82. Stress Due to a Centrifugal Force. — If a body of mass 
m, at the end of a cord, is whirled round in a circle the cord 
will be put in tension, the centrifugal force as we have learned 

from mechanics being equal to • where v is the linear 

velocity and r the distance of the center of gravity of the 
mass from the axis of rotation. If n is the number of revo- 
lutions per second, the angular velocity w will be equal to 
27rn. As the linear velocity is rw, the tension of the cord in 

terms of the angular velocity will be equal to 

Im being equal to ]. Consider a fly-wheel, the spokes of 

which represent a comparatively small part of the total weight, 
so that we may consider all the weight as being at the rim of 
the wheel. Let the thickness of the rim be t, and its distance 
from the axis of rotation be r. Let the weight of the material 
be IV lbs. per unit volume, and the breadth of the rim a 
units. Then the total weight of the rim will be w . 27rrta, the 
total radial force from the above formula is 

w . 27rrta X 4^Vr 
9 "~"^ 
and the radial force per unit area will be 

w . 27r7'ta X 4:7r-n-r w . 4:7r-n-rt 



R — 



g(27Tra) 



Strength of Material. 169 

This will give us the radial stress for a rim so thin that the 
stress may be considered uniform throughout the section. 
If, however, we consider a thick rim whose inside radius is 
?-2 and whose outside radius is i\ the increment of radial stress 
on a circular element whose thickness is di- will be given b}^ 
the above formula. We will have then for the increment of 
radial stress on this elemental volume 

,„ w.4:Tr^n^rdr _. 

dR=^ — . (1) 

The value of this increment having the negative sign because 
the radial stress decreases as the distance from the axis of 
rotation increases. To prove this latter statement suppose a 
bar of uniform section and length I rotates about an axis 
through its end. Let a be the sectional area, w the weight 
per unit volume, and x the distance of any point P from the 
axis of rotation. The stress at any point P is due to the 
centrifugal force of the part of the rod beyond P, therefore, 

as the center of gravity of this part is at a distance — - — 

from the axis and its weight is a(l — x)iv, the centrifugal 
force by the formula at the beginning of this article is 

2g ' 

It is clear from this equation that when x = I the C. F. is 
zero and when a; = it is a maximum. Proceeding then the 
integration of (1) gives 

C-^ being the constant of integration. We know the value oi 
the radial stress at the outer edge of the rim is zero, therefore 
when r = r^ 



170 Strength of Material. 

from which 



hencG;, 



This equation gives the radial stress at any distance r from 
the axis of rotation. 

N"ow considering the equilibrium of any particle we have 
by the same reasoning employed in Art. 80 the same two 
equations, 

H + R^C (3) 

and 

rdR + Rdr — Hdr. (4) 

Substituting in (4) the values of R and dR from (1) and 
(2) we have 

4:7rVivrdr , 4:7r^7i^w , ^ ^, , „, 
r . + —2g~ (^*i —r^)dr = Hdr. 

Integrating, the constant of integration being zero when 
r = we have 

ff = ^(V + r=). (5) 

Equations (2) and (5) give the radial and hoop tensile stress 
respectively for any point at a distance r from the axis of 
rotation, and both are independent of r2 . Therefore, these 
equations can be applied to solid wheels, such as a grindstone, 
as well as to fly-wheels. As in the preceding article these 
formulae have been deduced without considering the lateral 
deformation of Art. 79. The true hoop tension, taking this 
lateral deformation into consideration, "would be H-^:=H — IcR 
and the true radial stress R-^=^ R — TcH. The value of the 
radial stress is greatest where r ^ r^ (equation (2)), and for 
this value of r we have 



Steength or Material. 171 

^1 = " • ^"'"' {r,' - W - r,' - W) ■ (6) 

The hoop tension is greatest where r =^ r^ , hence 

From equations (6) and (7) it is clear that the hoop tension^ 
being the greater stress, will be the cause of rupture and also 
that the rupture will begin at the outside surface of the rim. 

83. Bending Due to Centrifugal Force. — We will have a 
bending moment due to centrifugal force if both ends of a 
straight rod are constrained to move in circles, for example, 
the horizontal rod between the two driving wheels of a loco- 
motive. In this case each point of the rod at any instant is 
moving in a circle and the centrifugal force due to this 
motion acts in the direction of the center of that circle for 
that instant. The stress caused acts in a diametrically oppo- 
site direction. Obviously the effect will be greatest when the 
horizontal rod is at its lowest point for here the weight of the 
rod itself acts down, as does also the stress due to centrifugal 
force. Calling the radius of the circle described by the ends 
of the rod r, and iv .the weight per unit length of the rod 
(considered of uniform cross-section) the radial stress due to 
centrifugal force will be 

gr 
The value of the linear velocity v is obtained from our 
knowledge of the speed of the engine, radius of the drive 
wheels, etc. For example, if the speed of the engine is 8, 
radius of the drive wheels i\ , and radius of motion of the 
ends of the rod r, the linear velocity of any point in the 

horizontal rod will be /S — , and this gives 

r, ^ 



172 Strength of Material. ^^^^^^H 

This value of R is uniform throughout tlie length of th^roT 
therefore^ the rod may be considered as a beam loaded uni- 
formly with R per unit length, and will have in addition 
when at its lowest position a uniform load due to its own 
weight. The stress at any point can then be readily found 

from the formula for bending ^ — _— . 

A connecting rod offers an important example of bending 
stress due to centrifugal force, but in this case one end of the 
rod is constrained to move in a straight line while the other 
end moves in a circle. At the instant the connecting rod is 
at right angles with the crank arm (obviously the greatest 
effect is produced when the load is perpendicular to the con- 
necting rod) if the end of the connecting rod is above the 
center of motion of the crank arm, the weight of the connect- 
ing rod will then act down, while the line of action of the 
stress due to the centrifugal force will act upward in a direc- 
tion parallel to the crank arm. Therefore, as before, the 
greater stress will be produced when the end of the connect- 
ing rod is lelow the center of motion of the crank arm. In 
this position, however, the force acting on the piston will put 
the connecting rod in tension so that it can readily sustain 
this bending stress. When the end of the connecting rod is 
above the center of motion of the crank arm, the stress pro- 
duced by the pressure on the piston is compressive and the 
connecting rod is in the condition of a strut and any bending 
due to centrifugal force will become an important matter 
(Chapter XI). Taking the position then as shown in Fig. 71. 
The radius of the circle described by any point of the con- 
necting rod will vary directly as the distance from A of the 
point, therefore, as the radial stress at any point due to the 

motion of the connecting rod is equal to — , where r is the 

radius of the circle described by the point and the value of /; 



Strexgth of Material. 173 

depends upon constants and the length of r, we must have 
the load on the connecting rod due to the centrifugal force 
vary uniformh' from zero at A to a maximum at B where it 

is equal to ^^^ where a is the length of the crank arm. The 

bending moment due to this load (Art. -^2), taking A for the 
origin, is 

/ being the length of the connecting rod, and x the distance 



Fig. 71. 

along it from A to am- point. This bending moment tends 
to bend the rod so that the middle of it will move upward. 
The weight of the rod itself acts vertically downward and the 
component of the weight which acts perpendicular to the rod 
will be IV cos Q per unit length. This load causes a bending 
moment equal to (Art. 38) 

the effect of which is to cause the middle of the rod to move 
11 



174 Steexgth of Material. 

downward. The total bending moment at any point distant 
X from A is then 

Having the bending moment at any point due to the motion 
and weight of the rod the stress due to these causes is readily 

found from ^ — _ . But it may be repeated that a con- 
necting rod in addition to the above stress suffers a com- 
pressive stress, when in this position, due to the pressure on 
the piston and also a bending stress due to this compressive 
load which puts the rod in the condition of the column or 
strut discussed in Chapter XI. Eeferring to that chapter it 
will be seen that the comparatively slight bending due to 
centrifugal force assumes important dimensions when we 
consider that the least variation of the axis of the connecting 
rod from the line of action of this large compressive load 
makes the opportunity for the load to cause bending. 

84. Flat Plates. — Experiment has proved that a circular 
flat plate when subjected to too great a pressure on one side 
always breaks along a diameter. Any diameter, then, of a 
circular plate is perpendicular to the greatest tensile stress 
due to bending caused by the pressure on one side of the plate. 
Let us consider a circular plate simply supported at the cir- 
cumference and subjected to a uniformly distributed load on 
the side opposite to the support. If the plate were fixed at 
the circumference it would be stronger than the one we are 
considering, just as a beam fixed at the ends is capable of 
supporting a greater load than if it were free at the ends. 
We will consider our support as a ring on which the plate 
rests, the pressure being uniformly distributed on the upper 
side. 

Let Fig. 72 represent such a plate, the support being at the 



Strength of Material. 



175 



circumference^ and the load iv lbs. per unit area. We must 
first find the bending moment at the diameter AB. The 

area of the triangular element shown is —. rdS, and the load 

on it is w-^dS. The supporting force under the arc rdO 

must be equal to this load^ and its distance from the diameter 
AB is r cos 6. The center of gravity of the uniform load on 




the triangular element is at a distance fr from the center 
and f r cos 6 from the diameter AB. The bending moment 
at the diameter AB due to these two forces is then 
_ lor"^ J A 9.. „„. /I I to 



dM = — ^de 

Integrating between the limits 



I /. cos ^ + ^ r^de . r cos 6. 



-^ and -^ for 6 
Z Z 



M = 



tur^ 



176 Strength of Material. 

which is the bending moment due to the loads on one side of 
the diameter AB. Letting t be the thickness of the plate 
the moment of inertia of the section through AB about the 

neutral axis is / = -^ . We have then from the equation of 
o 

bending ^ —-j.. {y ^^ this equation being equal to— .) 
y J. 4i 



wr^ t 6 



W-[^ 



which gives the stress on any section through a diameter. In 
designing a plate of this kind we would know the pressure to 
which it would be subjected and the dimensions of the open- 
ing it would cover, so if we put the limiting tensile strength 
of the material we are to use for p we can solve the above 
equation for t, the thickness necessary. 

If the load instead of being uniform is a concentrated one ; 
calling it W and supposing it is to be at the most effective 
position, the center, and to bear uniformly on a small circle 
of radius r^ concentric with the supporting ring, the part of 

W 

the load on one side of any diameter will be -^ and its center 

of gravity will be at a distance-^ from the diameter. The 

supporting force will be uniform throughout the length of 
the semicircle of the ring on the same side of the diameter as 
this load, and the center of gravity of this semicircle is at a 

distance — from the diameter and the total supporting force 

W 

on this semicircle is obviously -— . The bending moment at 

this diameter will be • 

W 2r_W Ar,_W( 2rA 

^*^- 2 * TT 2 • St: 7Z V '^ I' 

t being the thickness of the plate and the moment of inertia 



Stkength of Material. 177 

of the section be / = —^r- as before we have for the stress in 
o 

the section 

t 6 _3^/^ 2r^ 

Br 

If i\ =z r; or, which is the same thing, if the plate bears a 
uniform load, we have, remembering W will now be equal 

to Trr^LV, 

and if )\ ^= 0, or the load is concentrated at a point. 



My Wf 2rA t 6 SW/^ 



P = 






The stress given by these formulae is the maximum stress 
which occurs as one would expect at the center of the plate. 
Experiment proves this to be the case, for such plates begin 
to crack at the center and the crack extends from there, along 
a diameter, to the edge of the plate. Experiment also proves 
the above formulae to give very approximate results, so, 
though this may be a tentative method of deducing them (the 
assumptions not being strictly true), they will serve for all 
practical purposes. 

Rectangular Plates. — Experiment shows that rectangular 
plates when the length is not more than about twice the 
width, will crack along a diagonal. Therefore, if the sides 
are a and h (Fig. 73,), the diagonal section will support the 
greatest bending stress. Suppose the plate of thickness t to 
support a uniform load of iv per unit area. The load on one 

side of the diagonal will be^^^^ and its center of gravity will 

be at a distance -^ from the diagonal. Assuming the sup- 
porting force to be uniform along the edges, the part acting 
along the edge a will act at its center of gravity which is at 



178 



Strength of Material. 



a distance J from the diagonal, as is also the center of gravity^ 

of the part acting along the edge h. These two edges will 
support the whole load on this side of the diagonal, or the 



whole supporting force here will equal 
moment about the diagonal then is 

^_ wah X I wah x 



wab 



The bending 



wahx 
T2~ 



Calling g the length of the diagonal, the moment of inertia 




f - 



of the section through it about its neutral axis is, / 



gj 
12 



and 



The equation of bending gives us 

_ ^y _ 'i^cibx t 12 _ wabx 
~~r ~ T2" ' 2 ' Jt'~ 2g¥ ' 



Eef erring to the figure, ^ = Vc^2 + ^2 ^"^^ from similar 
ha 



triangles x ^ 



Va^ + 



, substituting these values 



P = 



wa-h'- 



If the plate is square a = & and we have 



p = 






Strength of Material. 179 

These formulae should . for practical purposes have a factor 
on the right member of the equation of 1.5 if the support is 
a fixed one (riveted joint), and of 5 if there is only a simple 
support. 

The formula for elliptical plates which experiment shows 
break along the major axis is for wrought iron and steel plates 
simply suported around the edges, 

(if cast iron is used the coefficient is 3 instead of f ) where tr 
is the load per unit area, t the thickness, and a and h the 
semi-major and semi-minor axes respectively. The theoret- 
ical solution for elliptical plates is very difficult because it 
involves elliptical integrals and because the supporting force 
is not uniform around the edge, as indeed is also the case 
for rectangular plates, but the variation for the latter is not 
excessive unless the plate is more than two or three times as 
long as it is wide. In fact all the above deductions are ap- 
proximations for the assumptions made are not strictly true. 
The formulae, however, agree closely with' the results obtained 
experimentally and may be assumed correct for all practical 
purposes. 

The subject of fiat plates is probably the most unsatis- 
factory one in the study of strength of material and practi- 
cal engineers have different methods for each form of plate. 
Fault may be found with the assumptions of most of these 
methods, though they all give approximate results. 

85. The principle of least work may be stated as follows: 
For stable equilibrium the stresses in any structure must have 
such values that the potential energy of the system is a mini- 
mum. The stresses of course must be within the elastic limit 
of the material. When forces act upon bodies which conform 



180 Strexgth of Mateeial. 

to Hook's law, the principle of least work may be applied to 
determine some unknown reaction. Perhaps the best-known 
example is that of a four-legged table which supports an un- 
symmetrically placed load. We can get from our knowledge 
of statics three equations to find the part of the load sup- 
ported by each leg, and the solution can be arrived at by 
means of the fourth condition furnished by the principle of 
least work. 

Distribution of Stress. — ^AVe have all along assumed that the 
stress is uniform throughout a section. As a matter of fact 
this is not the case. It can be mathematically proved that the 
shearing stress in a rod of square section varies as the ordi- 
nates of a parabola, being zero where the normal stress due 
to bending is a maximum, and a maximum where the normal 
stress due to bending is zero (at the neutral surface). Again, 
the shear parallel to the neutral axis in a rod of square sec- 
tion is zero, but in a rod of circular section it has a finite 
value. 

Velocity of Stress. — When a force is applied to a piece of 
material the stress is not instantaneously produced, but moves 
with a wave motion through the mass. The velocity of this 
motion can be found and it is shown to depend upon the 
stiffness and density of the material. The velocity of stress 
should be taken into account in problems involving impact 
and suddenly applied loads. 

Internal Friction. — When material is subjected to force 
and deformation occurs the molecules of the material move 
and this motion is resisted by internal friction. Heat is pro- 
duced and for the time between the application of the load and 
the complete rearrangement of the molecules, the stresses at 
planes through the material are changing; for this time then 
our formulee for planes of maximum stress are not correct. 



Strexgth of Material. 181 

Fatigue of Materials. — Experiment proves that material 
will break if subjected to repeated stress even if the stress be 
somewhat below the ultimate strength of the material. Ex- 
periment also shows that the greater the number of applica- 
tions the less becomes the value of the stress necessar}^ to cause 
rupture. For example, about a hundred thousand applica- 
tions of a stress of 49,000 lbs. to wrought iron will cause rup- 
ture, but if 500,000 applications were made the stress need 
be only 39,000 lbs. The loss of strength due to repeated stress 
is known as the Fatigue of Materials. This fatigue is more 
marked if the stress alternates from tension to compression 
and back again. 

The preceding facts have been mentioned to give the student 
the knowledge of their existence so that if inclined he may 
look them up in more complete works on the subject. The 
time limit of the course for which this book has been written 
precludes the possibility of entering into discussion of many 
subjects of importance, such as the stress in hooks, links, 
springs, rollers, foundations, arches and many others. 

Miscellaneous Examples : 

1. A reinforced concrete beam is 48 ins. wide, 54 ins. long 

and 5 ins. deep. It has a sectional area of 3.6 sq. in. of steel, 

the center of gravity of which is f in. below the top of the 

beam which is uniformly loaded with 2400 lbs. per in. length. 

-p 
(Use -^ = 10.) Find the position of the neutral surface, 

and the stress in the steel. 

Ans. Xeutral surface 3". 45 below top. 

'I. A reinforced concrete beam is 60 ins. long, 48 ins. wide 
and 5 ins. deep. The sectional area of the steel is 2.4 sq. in. 
and its center of gravity is -| in. below the top of the beam. 
The load is uniform and equals 3600 lbs. per in. length. 



182 Strej^tgth of Material. 

Find the position of the neutral surface and the stress in the 
concrete and steel. ^ = 10. 

Ans. Central surface 1".68 from top. 

3. If the allowable nnit stress in concrete be 500 lbs. per 
sq. in., and that for steel be -10,000 lbs. per sq. in., what per- 
centage of steel mnst be pnt in a beam 20 ins. wide by 10 
ins. deep if the steel is at a distance of 9 ins. from the top? 

Ans. 0.75^. 

4. If the allowable unit stress for concrete and steel are 
500 and 25,000 lbs. per sq. in. respectively, what is the resist- 
ing moment of a beam 7 ins. wide and 10 ins. deep if the 
reinforcement of steel is 1^ of the sectional area and placed 
at the lowest point of the beam? 

Ans. 92,800 in.-lbs. 

5. A beam is 8 ft. long and 1 ft. wide. The concrete is 
5 ins. thick and the steel reinforcement is 1 in. from the bot- 
tom of the beam. What area of steel section will be neces- 
sary, and using ^ in. square bars what should be the distance 
between them if a floor is made in this way ? 

6. The cross-section of a steel bar is 16 sq. in The bar 
is put under a stress of 27,000 lbs. per sq. in. li Tc ^ ^ 
what is the sectional area while under stress ? 

Ans. 15.99 sq. in. 

7. A steel bar is 2.5 in. in diameter and 18.5 ft. long. 
What are its- leng-th and sectional area under a pull of 
64,000 lbs. ? 

8. The inside of a cylinder is 6 ins., its outside radius is 
1 ft. The internal pressure is 600 lbs. per sq. in., and the 
external pressure is that due to the "atmosphere. What are 
the hoop and radial stresses at the outside and inside surfaces, 
also midway between these surfaces ? 

Ans. Hoop stress inside, 960; outside, 375 lbs. per sq. in. 



Strength of Material. 183 

9. A solid cylinder is under a uniform external pressure 
of 14,000 lbs. per sq. in. Show that the hoop and radial 
stresses are uniform and give values. 

10. A gun is built of a tube, inside radius 3 ins., outside 
radius 5 ins., and a jacket 2 ins. thick. Before assembling; 
the difference between the outside radius of the tube and the 
inside radius of the jacket was .004 in. What are the 
stresses at the outside and inside surfaces? 

Ans. Hoop compression, 14,400 lbs. per sq. in. 

11. In example 10 what powder pressure when firing the 
gun will just reverse the stress at the inner surface of the 
tube? 

12. The radii of a gun composed of tube and jacket are 
r3 = 3".04, r2i=5".8, and r^ = 9".75. The allowable unit 
stress is 50,000 lbs. per sq. in. for both tension and compres- 
sion. What are the radii before assembling? 

Ans. Eadius of bore, 3".0451; outside radius of tube, 
5".805 ; inner radius of jacket, 5".7915 ; outer radius, 9".7436 
ins. 

13. What internal powder pressure will stress the gun of 
example 12 to just 50,000 lbs. per sq. in. tension? 

Ans. 51,100 lbs. per sq. in. 

14. What is the greatest tangential stress in a cast-iron fly- 
wheel 30 ft. in diameter, rim 1 in. thick and 4 ins. wide, when 
it is making 60 revolutions per minute? 

Ans. Very roughly, 800 lbs. per sq. in. 

15. What is the stress in a cast-iron fly-wheel rim having a 
linear velocity of 1 mile a minute? 

Ans. Eoughly, 750 lbs. per sq. in. 

16. What diameter should a fly-wheel have if it is to make 
100 revolutions per minute, and the maximum allowable 
linear velocity is 6000 ft. per min. ? 

Ans. Eoughly, 19 ft. 



184 Strength of Material. 

17. A cast-iron bar 9 ft. long, 3 ins. wide and 2 ins. thick 
revolves about an axis J in. in diameter through it at a dis- 
tance of 4J ft. from one end. How many revolutions per 
second will produce rupture ? 

18. A solid steel circular saw is 4 ft. in diameter, and 
makes 2700 revolutions per minute. What is the stress at 
the circumference.? How many revolutions per minute will 
cause a stress of 35,000 lbs. ? 

19. An engine is making 750 revolutions per min., the 
connecting rod is 2 ft. long, and the crank arm 6 ins., ma- 
terial steel. What is the bending stress, due to centrifugal 
force, on the connecting rod if the area of its section be 1.5 
sq. ins. ? 

20. A cast-iron fly-wheel, mean diameter of rim 20 ft., 
makes 90 revolutions per minute. The cross-section of the 
rim is 10 sq. ins. What is the stress? 

21. What must be the thickness of a cast-iron cylinder 
head 36 ins. in diameter, allowable stress 3600 lbs. per sq. in., 
to sustain a load of 250 lbs. per sq. in. ? 

Ans. 5 ins. 

22. The allowable stress for steel being 12,000 lbs. per sq. 
in., how thick would a steel head for the cylinder of example 
21 have to be? 

Ans. 2.6 ins. 

23. A circular steel plate is 24 ins. in diameter and 1.5 ins. 
thick. It carries a load of 4000 lbs. at the center resting on 
a circle of 1 in. diameter. What is the maximum stress? 

Ans. 35,500 lbs. per sq. in. 

24. Suppose the load of example 23 were distributed on a 
surface of 3 ins. diameter. What would be the stress ? 

Ans. 10,100 lbs. per sq. in. 



Strength of Material. 185 

25. What must be the thickness of a steel plate 4 ft. square 
to carry 200 lbs. per sq. ft. 

Ans. 0.2 ins. 

26. What uniform load can be carried by a wrought-iron 
plate f in. thick^ 5 ft. long and 3 ft. wide ? 

Ans. About 12 lbs. per sq. in. 

27. A floor 18 ft. long, 15 ft. wide is made of concrete 
4 ins. thick, with 1 in. square wrought-iron rods, spaced 1 ft 
apart and at .75 in. from the bottom of the concrete. The 
floor carries a load of 150 lbs. per sq. ft. What is the maxi- 
mum stress? -^:= 15. 

Ans. Stress is about 450 lbs. per sq. in. 

28. An elliptical plate (cast iron) has a major axis 24 ins. 
long, a minor axis 16 ins. and is under a uniform pressure of 
22 lbs. per sq. in. The allowable stress is 3000 lbs. per sq. in., 
and the plate is simply supported at the edges. What must 
be the thickness? 

Ans. 1 in. 



i^OV 28 t&U8 



